Solved Problems

The work done by a constant force $\vec{F}$ over a displacement $\vec{r}$ is $W=\vec{F}.\vec{r}=F\left|r\right|{\cos  \theta\ }$ where $\theta$ is the angle between the force and displacement.  So 
\[W=\vec{F}.\Delta \vec{r}=F_x\Delta x+F_y\Delta y=\left(3\right)\left(2.5\right)+\left(-4.5\right)\left(-2\right)=16.5\, {\rm J}\] 

At the top, the normal force and gravity are in the same direction. These forces provide the centripetal force $F_r$ acting on the car. If the normal force $N$ becomes zero, then the car is on the verge of falling (but not yet falling). Apply Newton's second law and then set $N=0$. \begin{gather*} \Sigma F_r=ma_r\\\\ N_t+mg=\frac{mv^2_t}{r}\\\\ \Rightarrow\quad mg=\frac{mv^2_t}{r} \end{gather*} Substituting the numerical values into above, we will have \begin{align*}v_t&=\sqrt{Rg} \\ &=\sqrt{10\times 9.8} \\ &=9.9\quad \rm m/s \end{align*}

Use the conservation of mechanical energy between top and bottom points. \begin{gather*} U_{top}=mg(2R) \, , \, K_{top}=\frac{1}{2}mv^2_t \\\\ U_{bot}=0\, ,\,  K_{bot}=\frac{1}{2}mv^2_b \\\\ E_{top}=E_{bot} \\\\ \Rightarrow  2mgR+\frac{1}{2}m(gR)=\frac{1}{2}mv^2_b \\\\ \Rightarrow \, \boxed{v_b=\sqrt{5gR}=22.1\,\rm m/s} \end{gather*}

The forces act on the passengers at the bottom are shown in the figure. 
Note: in this problems the normal and centripetal forces always are toward the center of the circle i.e. $-\hat{r}$ and the gravity is in outward i.e. $\hat{r}$ direction.\begin{gather*} \Sigma F_r=ma_r \\\\  N(-\hat{r})+mg\hat{r}=\frac{mv^2_b}{r}(-\hat{r}) \\\\ \Rightarrow N_b=mg+\frac{mv^2_b}{r}=mg+\frac{5mgR}{R}=6mg \\\\ \therefore \quad \boxed{N=6mg=14.7\,\rm kN}\end{gather*}

By definition, the area under the force- displacement represents the work done by this force. So $W=\int{\vec {F}.d\vec {x}}$ or $W={\rm area(F-x)}$. Therefore, in this case we must find the area of a trapezoid with lengths of parallel sides $a$ , $b$ and height $h$.

Area of a trapezoid: $A=\frac{a+b}{2}.h$
So the work done by this force is 
\[W=\frac{(50{\rm kN}+20{\rm kN)}}{2}\left(400{\rm m}\right)=14\ {\rm MJ}\]  

Conservation of mechanical energy gives the velocity of the ball before the collision:
\[E_i=E_f\Rightarrow \ mg\,l\,\left(1-{\cos  \theta\ }\right)=\frac{1}{2}mv^2_1\] 
$\Rightarrow \ v_1=\sqrt{2gl\,\left(1-{\cos  \theta\ }\right)}$

Because of elastic collision, kinetic energy and momentum are conserved.

let $m_1$be the ball and $m_2$ be the block

Use the conservation of momentum $P_i=P_f$ to determine the velocity of the ball and block after the collision:
\[\Rightarrow m_1v_1=m_1v_{1f}+m_2v_2\to v_1=v_f+\frac{m_2}{m_1}v_2\ \ \ \ \ ,\ \ {\rm (I)}\] 

Conservation of K.E gives:  $\frac{1}{2}m_1v^2_1=\frac{1}{2}m_1v^2_{1f}+\frac{1}{2}m_2v^2_2\to v^2_1=v^2_{1f}+\frac{m_2}{m_1}v^2_2$

From ${\rm (I)}$ we have: $v^2_{1f}=v^2_1-2\frac{m_2}{m_1}v_1v_2+{\left(\frac{m_2}{m_1}v_2\right)}^2$
\[\therefore v^2_1=v_1-2\frac{m_2}{m_1}v_1v_2+{\left(\frac{m_2}{m_1}\right)}^2v^2_2+\frac{m_2}{m_1}v^2_2\] 
If the relation above divided by $\frac{m_2}{m_1}v_2$ we obtain: 
\[0=-2v_1+\frac{m_2}{m_1}v_2+v_2\ \ \ \Longrightarrow \ \ v_2=\frac{2v_1}{1+\frac{m_2}{m_1}}\] 
Now write the conservation of mechanical energy of spring:
\[\frac{1}{2}m_2v^2_2=\frac{1}{2}k{\left(\Delta x\right)}^2\ \ \Longrightarrow v_2=\sqrt{\frac{k}{m}}\Delta x=\sqrt{\frac{2000}{1}}\left(0.2\right)=0.894\frac{{\rm m}}{{\rm s}}\] 

by substituting this into above relation for $v_2$ and solve for $v_1$, we obtain
\[v_1=\frac{1}{2}v_2\left(1+\frac{m_2}{m_1}\right)=2.68\frac{{\rm m}}{{\rm s}}\] 
\[{\cos  \theta\ }=1-\frac{v^2_1}{2gl}\to \theta=50.7{}^\circ \] 

Using the conservation of mechanical energy between the initial and final points, we obtain \begin{gather*} K_i+U_i+W_{wind}=K_f+U_f \\\\ 0+mg(\underbrace{100\,\sin 10^\circ}_{h_i})-50\,{\cos 10^\circ}\times 100=\frac{1}{2}mv^2_f+0 \end{gather*} where $U_i$ is the initial potential energy of the skier. At the top of the slop, its height relative to the ground is $h_i=100\, sin 10^\circ$. $W_{wind}$ is the work done by the force of the wind on the skier. therefore, \begin{gather*} v^2_f=100\left(2g{\sin 10^\circ}-\frac{2\times 50\times {\cos 10^\circ}}{m}\right) \\\\ =100\left(2\times 9.8\times {\sin 10^\circ}-\frac{2\times 50\times {\cos 10^\circ}}{70}\right) \\\\ \Rightarrow\quad \boxed{v_f=14\quad \rm m/s}\end{gather*}

Use the conservation of the momentum to find the velocity of the bullet after the collision.
\[P_i=P_f\to mv=Mv_f+m\frac{v}{2}\ \Rightarrow v_f=\frac{m}{M}\frac{v}{2}\ \ \ (*)\] 
Because there is no frictional effects, so mechanical energy of bob is conserved.
\[E_i=E_f\to \ U_i+K_i=U_f+K_f\] 
Barely swing through a complete vertical circle means that the bob has some energy to reach only the top of the path. So use the conservation of the mechanical energy between bottom and top point of the circle. Let the lowest point of the pendulum as reference point $U_i(bottom)=0$ and $U_f(top)=Mg(2l)$
\begin{gather*} \frac{1}{2}Mv^2_f=Mg(2l)\to v_f=2\sqrt{gl} \\\\ \text{from} (*)  \Rightarrow \boxed{v_{min}=\frac{4M}{m}\sqrt{gl}}\end{gather*}

Recall that the total mechanical energy in SHM is $E=\frac{1}{2}mv^2+\frac{1}{2}kx^2=\frac{1}{2}kA^2$.
Where $A$ is the amplitude or maximum distance to the equilibrium. Therefore, before the collision, we have
\[E_1=U_1+K_1=\frac{1}{2}mv^2_1+0\left(in\ equilibrium\ position\right)=\frac{1}{2}kA^2_1\] 
\[\Rightarrow v_1=\sqrt{\frac{k}{m}}A_1\] 
Conservation of momentum states that $P_i=P_f\ \Rightarrow Mv_1=\left(M+m\right)v_2$
\[\Rightarrow v_2=\frac{M}{M+m}v_1\ \ ,\ after\ collision\] 
So the total mechanical energy of the system after the collision is 
\[E_2=U_2+K_2\ \Rightarrow \frac{1}{2}kA^2_2=0+\frac{1}{2}\left(m+M\right)v^2_2=\frac{1}{2}\frac{M^2}{M+m}v^2_1\] 
\[\Rightarrow E_2=\frac{M}{M+m}E_1\] 
\[\Rightarrow \frac{1}{2}kA^2_2=\frac{M}{M+m}\frac{1}{2}kA^2_1\to A_2=\sqrt{\frac{M}{M+m}}A_1\] 
In SHM, period is given by $T=2p\sqrt{\frac{m}{k}}$ so $T_f=2p\sqrt{\frac{M_{tot}}{k}}=2p\sqrt{\frac{m+M}{k}}\ \ $ 

At one ends of a SHM $v=0$  and $E_{tot}=\frac{1}{2}kA^2$ so
\[K_1=K_2=0\ \ ,\ E_{tot,1}=E^{'}_{tot,2}\Rightarrow \ \frac{1}{2}kA_1=\frac{1}{2}kA^{'}_2\ \ \ \] 
\[\therefore A_1=A^{'}_2\ {\rm ,\ does\ not\ change!}\] 
The period of motion does not change.
 

The collision between Jane and Tarzan is considered a perfectly inelastic collision. We can use conservation of energy to find Jane's speed just before the collision, and then again to find how high they go after the collision.

Stage 1: Jane swings
\begin{gather*} E_{i,mech}=U+K=m_{J}gL+0\\ E_{f,mech}=U_f+K_f=0+\frac{1}{2}m_{J}v^2_{J} \end{gather*} Apply conservation of mechanical energy as below \[E_i=E_f\to m_{J}gL=\frac{1}{2}m_{J} v^2_J\Rightarrow v_J={\left(2gL\right)}^{\frac{1}{2}}\]
Stage 2: Jane collide with Tarzan. Use conservation of momentum to find the combined velocity of Tarzan and Jane just after collision:
\begin{gather*} P_i=P_f \\\\ m_Jv_J=\left(m_T+m_J\right)V_{TJ}\\\\ \Rightarrow V_{TZ}=\frac{m_J}{m_J+m_T}v_J\end{gather*} Therefore,\[V_{TJ}=\frac{m_J}{m_J+m_T}{\left(2gL\right)}^{\frac{1}{2}}\] 
Stage 3: now use again the conservation of the mechanical energy to determine the desired height:
\begin{gather*}E_i=U_i+K_i=0+\frac{1}{2}\left(m_T+m_J\right)V^2_{TJ}\\\\=\frac{1}{2}\left(m_T+m_J\right){\left(\frac{m_J}{m_J+m_T}\right)}^22gL \\\\E_f=U_f+K_f=\left(m_T+m_J\right)gH+0 \\\\ E_i=E_f\Rightarrow H={\left(\frac{m_T}{m_T+m_J}\right)}^2L={\left(\frac{54}{136}\right)}^225=3.9{\rm m}\end{gather*}
Note that in above $U_i=0$ because in this problem we chose the lowest point of the path as a reference point. 

Working backward, if we know the horizontal velocities at the end of the ramp, the distances traveled are $L=V\Delta t$ and $L^{'}=V^{'}\Delta t$ where $\Delta t$ is the time for both to fall vertically from the bottom of the ramp to the ground. 

Thus, we conclude $\frac{L}{L'}=\frac{V}{V'}$.
Using the conservation of mechanical energy, we can find $V$ and $V'$. (Let the end of the ramp to be as the reference point i.e. $U=0$)
\begin{gather*} E_i=mgH\\E_f=K_f+U_f=\frac{1}{2}I\omega^2+\frac{1}{2}mV^2\end{gather*}Recall that for a spherical shell, $I=\frac{2}{3}mR^2$, and for a solid sphere, $I'=\frac{2}{5}mR^2$. Thus, the conservation of mechanical energy for spherical shell is as below: \begin{align*}E_i&=E_f \\\\ mgH&=\frac{1}{2}\left(\frac{2}{3}\right)mR^2{\left(\frac{V}{R}\right)}^2+\frac{1}{2}mV^2\\\\&=\frac{5}{6}mV^2 \\\\ \Rightarrow \quad  V&=\sqrt{\frac{6}{5}gH}\end{align*} 
Similarly, the conservation of mechanical energy for solid sphere is written as below \begin{align*} E_i&=E_f\\\\ mgH&=\frac{1}{2}\left(\frac{2}{5}\right)mR^2{\left(\frac{V'}{R}\right)}^2+\frac{1}{2}m{V'}^2\\&=\frac{7}{10}m{V^{'}}^2\\\\ \Rightarrow\quad  V'&=\sqrt{\frac{10}{7}gH}\end{align*} Therefore, the ratio between the two distances are as follows \begin{align*} \frac{L}{L'}&=\frac{V}{V'}\\\\&=\frac{\sqrt{\frac{6}{5}gH}}{\sqrt{\frac{10}{7}gH}}\\\\&=\sqrt{\frac{42}{50}}\\\\ \Rightarrow \quad \frac{L'}{L}&=1.09\end{align*}

(a) Work-energy theorem states that $\Delta K=k_f-K_i=W_{net}$, where $W_{net}$ is the total work done on the system. Work is related to force by the formula  $W=\int{\vec{F}\cdot d\vec{x}}$
\begin{align*}W=\int^L_0{\vec{F}\cdot d\vec{x}}&=\int^L_0{F\,dx\,{\cos 0^\circ}}\\&=\int^L_0{bx^2dx}\\&=\frac{1}{3}b{\left.x^3\right|}^L_0\\&=\frac{1}{3}bL^3\end{align*} Particle starts at rest, so $K_i=0$.
\[K_f-K_i=W\Rightarrow K_f=\frac{bL^3}{3}\] 
(b) Solving $K_f$ for speed, we obtain
\begin{gather*} K_f=\frac{1}{2}mV^2_f=\frac{bL^3}{3} \\\\ \Rightarrow \quad V_f=\sqrt{\frac{2}{3}\frac{bL^3}{m}}\end{gather*}

(a) Because the tension along the path is constant and parallel to the direction of motion so the physical work done by it is given by \[W_T=\int^x_0{\vec{F}.d\vec{l}}=\int^x_0{T\,dl}=Tx\]

For more problems on work in physics, refer to the following:
Work problems in physics

(b) Use the work-energy theorem to find the speed of the box.
\[K_f-K_i=W_{tot}=W_{weight}+W_{Tension}\] 
\[W_{weight}=\int^x_0{{\vec{F}}_g \cdot d\vec{l}}= \begin{cases} \int^x_0{\left(mg\,{\sin  \theta }\right)dx\, \cos  180^\circ} &=-mg\,x\,{\sin \theta } \\ \\ \int^x_0{\left(mg\,{\cos \theta }\right)dx\,\cos  90^\circ} & =0 \end{cases}\] The box starts at rest, $K_i=0$ so \begin{gather*}\frac{1}{2}mV^2_f=-mg\,x\,{\sin \theta\ }+Tx\\ \\ \Rightarrow V_f={\left(\frac{2Tx}{m}-2gx\,{\sin  \theta\ }\right)}^{\frac{1}{2}}\end{gather*}

(c) Power is defined as the rate of the work done by the system \begin{gather*} P=\frac{dW}{dt}=\frac{d}{dt} \vec{F}\cdot\vec{x} \\\\ \xrightarrow{F\ is\ constant} P=\vec{F}\cdot\frac{dx}{dt}=\vec{F}\cdot\vec{x}=\vec{F}\cdot\vec{V} \end{gather*} 
In this problem, $\vec{T}$ and $\vec{V}$ are parallel (weight is normal to velocity), so \[P=TV=T{\left(\frac{2Tx}{m}-2gx\,{\sin \theta}\right)}^\frac 12\] 

The free-body diagram of an object sliding on a ramp is as follows:

There is no movement in the perpendicular direction to the motion, i.e., 
\[\Sigma F_y=0\to N-mg\,{\cos  \theta\ }=0\ \ ,\ \ (1)\] 
On the other hand, along the inclined plane $\Sigma F_x=0$, since the block is at the threshold of the moving.  \[\Rightarrow T-f-mg\,{\sin  \theta\ }=0 \quad ,\quad (2)\]  At the instant, the block begins to move, static friction has reached its maximum value, i.e. $f_{s,max}=\mu_{s}N=\mu_{s}mg\,{\cos \theta}$, thus, substituting this into $(2)$, we will have \[(2) \Rightarrow \ T=\mu_s mg\,{\cos \theta}+mg\,{\sin \theta}\]

If the spring is not accelerating, then the force pulling down must be equal to the tension force on the rope over the pulley so, $F_{spring}=kx$, where $x$ is the distance the spring is stretched. Therefore \begin{gather*} T=\mu_s mg\cos\theta+mg\sin\theta \\ T=kx \end{gather*} Equating these two expressions, we will have \[x=\frac{mg( \mu_s \cos\theta+\sin\theta)}{k}\]

The potential energy stored in the spring is calculated as follows \begin{align*} U_{spring}&=\frac{1}{2}kx^2 \\\\ &=\frac{1}{2k}[mg\left(\sin \theta+\mu_{s}\,\cos \theta)\right]^2\end{align*}

To solve more problems on potential energy, refer here. 

(a) Use conservation of mechanical energy in the presence of dissipating forces like friction. i.e. $\Delta E=E_f-E_i=W_f$, where $W_f$ is the work done by these forces.
Initially the box is at rest, $K_i=0$ and the spring is stretched, $U_{i,s}=\frac{1}{2}kd^2_0$. 
When it comes to a stop, $K_f=0$ again and the final potential energy of the spring is $U_{f,s}=\frac{1}{2}k{\left(\Delta x\right)}^2$, where $\Delta x=d_1-d_0$ is the final displacement of the spring from equilibrium position. 
Kinetic friction has done work over the block in this path so
\[W_{friction}=-f{{\rm d}}_{{\rm 1}}=-\mu_{k}mg\,d_1\] 
Therefore: 
\[E_f-E_i=W_f\to \left(U_f+K_f\right)-\left(U_i+K_i\right)=W_f\] 
\[\frac{1}{2}k{\left(d_1-d_0\right)}^2-\frac{1}{2}kd^2_0=-\mu_{k}mg\,d_1\] 
\[d^2_0-d^2_1+2d_1d_0-d^2_0=2\mu_{k}\,mg\,d_1\Rightarrow d_1=2d_0-2\mu_{k}\frac{mg}{k}\] 
(b) Use conservation of mechanical energy, $\Delta E=W_f$
When the box reaches $d_0$ there is no spring potential energy (the box is on the equilibrium position) i.e. $U_{f,s}=\frac{1}{2}kd^2_0$ but at this point the box has velocity i.e. $K_f=\frac{1}{2}mv_{f2}$
\[E_f-E_i=W_f\Rightarrow \left(0+\frac{1}{2}mv^2_f\right)-\left(\frac{1}{2}kd^2_0+0\right)=-\mu_{k}mg\,d_0\] 
\[v_f={\left(2\frac{k}{m}d^2_0-2\mu_kg\,d_0\right)}^{\frac{1}{2}}\] 
\[\ \] 
(c) Setting $d_1=d_0$ in the result of part (a) and solving for $\mu_k$, we obtain
$$d_1=2d_0-2\mu_k\frac{mg}{k}\ \ \xrightarrow{d_0=d_1} \mu_k=\frac{kd_0}{2mg}\ $$

(a) Because at this point the system is at rest so \begin{gather*}\Sigma F_x=0\Rightarrow T{\cos 36.9^\circ}-kx=0\\\\ \Rightarrow T=\frac{kx}{{\cos 36.9{}^\circ \ }}=\frac{120\times 0.40}{{\cos 36.9{}^\circ \ }}=60.02\ {\rm N}\end{gather*}

(b) This part is related to the work-energy theorem problem. According to this theorem, we have $K_f-K_i=W_{net}$. Because the block is initially at rest and after $x=0.4\ {\rm m}$ again come to a stop, so $K_i=K_f=0$. 

\[\Delta K=W_{net}\Rightarrow 0-0=W_T+W_s\] 

Where $W_s=-\frac{1}{2}kx^2$ is the work done by the spring on the block and $x$ is distance to equilibrium. Therefore, \[W_T=\frac{1}{2}kx^2=\frac{1}{2}(120)(0.4)^2=9.6\quad {\rm J}\]

Important note: we cannot use the definition of the work done by a constant force $F$ on an object over a distance $x$ that is $W=\vec{F}\cdot d\vec{x}$, because, in this case, block attached to spring and so the tension in the rope is not a constant force so $W\ne \vec{T}\cdot d\vec{x}$. 

(c) Use the conservation of mechanical energy between the instant that the rope breaks and when it is reached an equilibrium. \begin{gather*} E_2-E_1=W_{NC}=0 \quad \Rightarrow\quad E_2=E_1\\\\ K_2+U_2=K_1+U_1 \\\\ \frac{1}{2}mv^2_2+0=0+\frac{1}{2}kx^2 \\\\ \Rightarrow v=\sqrt{\frac{k}{m}}x=\sqrt{\frac{120}{5}}\left(0.4\right)=1.96\,{\rm \frac ms} \end{gather*}

In the above, $W_{NC}$ is the work done by the non-conservative forces such as friction.

Note: the potential energy stored in the spring is the negative of the work done by it that is \[U_s=-W_s=\frac{1}{2}kx^2\] 

Because there is a non-conservative force acting on the $6\, {\rm kg}$ block so we must use the following version of conservation of the mechanical energy:
\[E_2-E_1=W_{NC}\] 

Let the initial position of the blocks be the origin of the coordinate systems, so the $2\, {\rm kg}$ block moves downward the coordinate. Conservation of energy gives,
\begin{gather*} E_f-E_i=W_f\\ \to \left(U_f+K_f\right)-\left(U_i+K_i\right)=-fd \end{gather*}
Where $d$ is the distance that the $6\ {\rm kg}$ block moves (or both blocks moves!)
\begin{gather*}\left(-m_2gd+\frac{1}{2}\left(m_6+m_2\right)v^2_f\right)-\left(0+0\right)=-\mu_km_6gd\\\Rightarrow \frac{1}{2}\left(m_6+m_2\right)v^2_f=gd\left(m_2-\mu_km_6\right)\\\end{gather*} Therefore, the final speed is obtained by solving the above equation for $v_f$
\begin{align*} v_f&=\sqrt{\frac{2gd\left(m_2-\mu_km_6\right)}{m_6+m_2}}\\\\&=\sqrt{\frac{2\times 9.8\times 0.30\times (2-0.15\times 6)}{6+2}}\\\\&=0.899\quad {\rm \frac ms} \end{align*}
Note: the friction work is always $W_f=-fd=\mu_kNd$, where $N$ is the normal force and $d$ is the displacement of the object.
Note:  the $2\, {\rm kg}$ object is moving down the origin so its potential energy must be negative.

Use the work-kinetic energy theorem: $\Delta K=K_f-K_i=W_{net}$, where $W_{net}$ is the sum of works done on the object in the displacement $x$. Therefore,
\begin{align*}\frac{1}{2}m(v^2_f-v^2_i)&=Fx \\\\ \Rightarrow \quad v_f&=\sqrt{\frac{2Fx}{m}} \\\\ &=\sqrt{\frac{2\times 95\times 0.80}{0.080}} \\\\ &=44\quad \rm m/s \end{align*}

By definition, the work done by a constant force $\vec{F}$ over a distance $\vec{x}$ is scalar product of vector force by the displacement vector $W=\vec{F}\cdot \vec{x}$ so \begin{align*} W&=(12\,\hat{i}-10\,\hat{j})\cdot (12\,\hat{i}+11\,\hat{j})\\ &=144(\hat{i}\cdot \hat{i})-110(\hat{j}\cdot \hat{j})\\ &=144-110\\ &=34\quad {\rm J} \end{align*} Note: the dot product or inner product of the unit vectors are defined as \[\hat{i}\cdot \hat{i}=|\hat{i}||\hat{i}|{\cos 0{}^\circ}=1\] And so on \[\hat{j}\cdot \hat{j}=\hat{k}\cdot\hat{k}=1\] And similarly, \[\hat{i}\cdot\hat{j}=|\hat{i}||\hat{j}|{\cos 90{}^\circ}=0\] 

Use the work-energy theorem $\Delta K=W_{net}$, where $W_{net}$ is the total work done on an object. In this case, there are two works, due to the constant force and friction. First calculate the work $W_P$ and substitute it in the $\Delta K=W_{net}$, then solve for $W_f$.
\[W_P=P\left|x\right|{\cos  \theta\ }=160\left(8\right){\cos  30{}^\circ \ }=1108.51\ {\rm J}\] 
\[K_f-K_i=W_f+W_P\Rightarrow \frac{1}{2}m\left(v^2_f-v^2_i\right)=W_f+W_P\] 
\[\frac{1}{2}\left(20\right)\left({\left(2.6\right)}^2-{\left(0.5\right)}^2\right)=W_f+1108.51\Rightarrow W_f=-1043.41\ {\rm J}\] 
 

Use the work-energy theorem but in this case, we have a varying force so 
\[K_f-K_i=W_{net}=\int^{x_f}_{x_i}{F\left(x\right)dx}\] 
\[\Rightarrow \frac{1}{2}m\left(v^2_f-v^2_i\right)=\int^{x_f}_{x_i=0}{\left(-C+Dx^2\right)dx}={\left(-Cx+\frac{D}{3}x^3\right)}^{x_f}_{x_i=0}\] 
\[\Rightarrow v_f=\sqrt{v^2_0+\frac{2}{m}\left(-Cx_f+\frac{D}{3}x^3_f\right)}\ \ \] 
 

(a) There are two forces acting on the boy, one is the normal force and the other is gravity as shown in the figure below.

(b) Use the conservation of mechanical energy between points $A$ and $P$. Consider the center of the hemisphere as the reference level so at $P$ the distance to the base is $h=R\, \cos \theta$
\[E_A=E_P\Rightarrow U_A+K_A=U_P+K_P\] 
\[\Rightarrow \ \ mgR=mgR\,{\cos \theta\ }+\frac{1}{2}mv^2_P\ \] 
\[\Rightarrow v_P=\sqrt{2gR(1-{\cos \theta\ })}\] 
This is the velocity of the boy everywhere on the hemisphere.

(c) When the normal force acting on the boy reaches zero, then the boy flies off. So first apply Newton's 2${}^{nd}$ law on the boy at point $P$, then set $N=0$:
\[\Sigma F_r=ma_r\Rightarrow N\hat{r}+mg\,{\cos \theta\ }\left(-\hat{r}\right)=\frac{mv^2}{R}\left(-\hat{r}\right)\] 
\[\Rightarrow mg\,{\cos \theta\ }-N=\frac{mv^2}{R}\xrightarrow{N=0} v=\sqrt{Rg\,{\cos  \theta\ }}\] 
Note: we have used the polar coordinate. The centripetal acceleration is always towards the center i.e. $-\hat{r}$ 

In part (b) we have found the velocity of the boy at an arbitrary angle $\theta$. In part (c) the velocity at instant the boy flies off have found. Now set equal these velocities with each other to determine the angle at which the boy files off.
\[\sqrt{2gR\,\left(1-{\cos  \theta\ }\right)}=\sqrt{Rg\,{\cos \theta\ }}\Rightarrow 2gR\,\left(1-{\cos  \theta\ }\right)=Rg\,{\cos \theta\ }\] 
\[\Rightarrow 2Rg=3Rg\,{\cos \theta\ }\Rightarrow \theta={{\cos }^{-1} \left(\frac{2}{3}\right)\ }\ \sim \ 48.19{}^\circ \] 
 

Use the conservation of mechanical energy as $E_i=E_f$.

Here there is a rotational kinetic energy due to the rolling of the sphere, therefore

\[\frac{1}{2}mv^2_i+\frac{1}{2}I\omega^2+mgh=\frac{1}{2}mv^2_f+\frac{1}{2}I\omega^2+0\] 
Where $I$ is the moment of inertial that for a sphere is $\frac{2}{5}mR^2$. Recall that in the rotational motion the angular velocity is related to the linear velocity via $v=r\omega$ so 
\[\frac{1}{2}mv^2_i+\frac{1}{2}\left(\frac{2}{5}mR^2\right){\left(\frac{v_i}{R}\right)}^2+0=\frac{1}{2}mv^2_f+\frac{1}{2}\left(\frac{2}{5}mR^2\right){\left(\frac{v_f}{R}\right)}^2++mgd\,{\sin  25{}^\circ \ }\] 

\[\frac{7}{10}mv^2_i=\frac{7}{10}mv^2_f+mgd\,{\sin  25{}^\circ \ }\Rightarrow v_f=\sqrt{v^2_i-\frac{10}{7}gd\,{\sin  25{}^\circ \ }}\] 

\[\Rightarrow v_f=\sqrt{{\left(4.5\right)}^2-\frac{10}{7}\left(9.8\right)\left(3\right){\sin  25{}^\circ \ }}=1.58\frac{{\rm m}}{{\rm s}}\] 
 

Use the conservation of mechanical energy to find the desired speed.
\[E_1=E_2\Rightarrow \frac{1}{2}mv^2_1-\frac{GmM}{r_1}=\frac{1}{2}mv^2_2-\frac{GMm}{r_2}\] 
Where the second term is the gravitational potential of an object of mass $m$ at a distance $r\ $from the center of a planet with mass $M$. 
\[v^2_1-\frac{2GM}{r_1}=v^2_2-\frac{2GM}{r_2}\Rightarrow v_2=\sqrt{v^2_1+2GM\left(\frac{1}{R+h}-\frac{1}{R}\right)}\] 
\[\Rightarrow v_2=\sqrt{{\left(2000\right)}^2+2\left(6.67\times {10}^{-11}\right)\left(3.95\times {10}^{23}\right)\left(\frac{1}{\left(5\times {10}^6+{10}^6\right)}-\frac{1}{5\times {10}^6}\right)}=1498\frac{{\rm m}}{{\rm s}}\] 

Note: the gravitational potential energy is equal to the work done against the gravity of an object with mass $M$ to bring a unit mass $m$ to a given point say $r$ from the center of it .so 
\[U=-\frac{GmM}{r}\] 
Where $G=6.67\times {10}^{-11}\, \rm N\cdot m^2/kg^2$ is the gravitational constant. 

(a) Use the conservation of the mechanical energy between starting point $i$ and top of the loop $t$. (let the lowest point of the circle as the base)
\[E_i=E_t\Rightarrow U_i+K_i=U_t+K_t\Rightarrow \ mgh+0=mg\left(2R\right)+\frac{1}{2}mv^2_t\] 
\[v_t=\sqrt{2g(h-2R)}=\sqrt{2\left(9.8\right)\left(50-2\left(15\right)\right)}=19.79\ {\rm m/s}\] 

(b) There is a centripetal force acting on the cart provided by the normal force in the circular path. Recall that this force always is in direction of the center of the circle ($-\hat{r}$). So apply Newton's 2${}^{nd}$ law to the cart
\[\Sigma {\vec{F}}_r=m{\vec{a}}_r\Rightarrow N\left(-\hat{r}\right)+mg\left(-\hat{r}\right)=\frac{mv^2}{2}\left(-\hat{r}\right)\] 
\[N=\frac{mv^2}{r}-mg\] 
Thus the normal force is, in general, as above. Put the velocity of the top point of the loop to find the normal force at that point
\[N=\frac{mv^2_t}{r}-mg=\left(1.2\times {10}^3\right)\left(\frac{{\left(19.79\right)}^2}{15}-9.8\right)\ \sim \ 19.6\ {\rm kN}\] 
 

(a) Rolls without slipping means that the point of the contact of the object with the ground does not move. This type of motion is provided by static friction. Therefore, three forces act on the sphere. Gravity, normal force, and static friction.
(b) The static force does not any work on the sphere since the sphere instantaneously does not slide but rolls! Therefore, use the conservation of mechanical energy to find the desired height.
\[E_f-E_i=W_{NC}=0\Rightarrow E_f=E_i\] 
\[mgh=\frac{1}{2}mv^2_A+\frac{1}{2}I\omega^2+mg\left(2R\right)\] 
If substitute $I=\frac{2}{5}mr^2$ and $\omega=v/r$ then we obtain
\[mgh=\frac{1}{2}mv^2_A+\frac{1}{2}\left(\frac{2}{5}mr^2\right){\left(\frac{v_A}{r}\right)}^2+mg\left(2R\right)\Rightarrow gh=\frac{7}{10}v^2_A+2gR\] 
Now apply Newton's 2${}^{nd}$ law in the circular motion at the point A to find the $v_A$
\[\Sigma F_r=ma_r\Rightarrow N+mg=\frac{mv^2_A}{R}\ \] 
If we set $N=0$, we find the minimum value of $v_A$ so $v^{min}_A=\sqrt{Rg}$. Put this value into the relation above 
\[gh=\frac{7}{10}\left(Rg\right)+2gR\Rightarrow \ \ h_{min}=2.7R\] 
 

(a) By definition, the work done by a constant force $\vec{F}$ is the product of the magnitude $\Delta \vec x$ of the displacement of the point of application of the force and the component of the force along the direction of the displacement that is $F{\cos  \theta\ }$. Therefore 
\[W=\vec F.\Delta \vec x=F\left|\Delta x\right|{\cos \theta\ }\] 
First, draw all of the forces act on a body on an incline plane as shown in the figure. Recall that the components along the direction of the displacement does work and the components perpendicular to the displacement does not work (since the angle between force and displacement is $\theta=90{}^\circ $)
\[W_g=\left(\underbrace{mg\,{\sin  40{}^\circ \ }}_{F_{\parallel }}\right)\times 1.6\times {\cos  0{}^\circ \ }=8\times 9.8\times {\sin  40{}^\circ \ }\times 1.6=+80\ {\rm J}\] 
\[W_N=Nx\,{\cos  90{}^\circ \ }=0\] 
\[W_f=fx\,{\cos  180{}^\circ \ }=-fx=-\mu_kNx\] 
Since the coefficient of kinetic friction have not given so we must find the work done by the friction by applying the work-energy theorem as follows 
\[\Delta K=W_{tot}=W_g+W_N+W_f\] 
\[\frac{1}{2}m\left(V^2_2-V^2_1\right)=W_g+W_f+W_N\Rightarrow \frac{1}{2}\left(8\right)\left(4^2-0^2\right)=80+W_f+0\] 
\[W_f=-16\ {\rm J}\] 

(b) The time rate at which a force does work is called Power. Therefore, calculate the change in the kinetic energy of the block then use the definition of the power to find it
\[P=\frac{energy}{time}\] 
\[P=\frac{\Delta K}{\Delta t}=\frac{\frac{1}{2}m\left(V^2_2-V^2_1\right)}{\Delta t}=\frac{\frac{1}{2}\left(4\right)\left(4^2-0^2\right)}{0.8}=80\ {\rm W}\] 
 

Initially, the spring is compressed so it has elastic potential energy. After shooting out all of its potential energy converts to kinetic energy. Using the conservation of mechanical, we obtain
\[E_f-E_i=W_{NC}\] 
Where $W_{NC}$ is the work done by the non-conservative forces such as friction. In this case, $W_f=0$ so we have
\[E_f=E_i\] 
\[\Rightarrow \underbrace{\frac{1}{2}kx^2_f}_{elastic\ potential}+\underbrace{\frac{1}{2}mv^2_f}_{kinetic\ energy}=\frac{1}{2}kx^2_i+\frac{1}{2}mv^2_i\] 
When the spring reaches its unstretched length ($x=0$), the ball is shot. Therefore, 
\[0+\frac{1}{2}mv^2_f=\frac{1}{2}kx^2+0\Rightarrow v_f=x\sqrt{\frac{k}{m}}=\left(0.05\right)\sqrt{\frac{10}{0.02}}=1.12\,\rm m/s\] 

(a) consider $h$ to be the height of the child from releasing point to the lowest point. From the sketch, we obtain 
\[\ell\,cos\ 45{}^\circ +h=\ell \Rightarrow h=\ell\left(1-{\cos  45{}^\circ \ }\right)\] 
\[\Rightarrow h=3\left(1-\frac{\sqrt{2}}{2}\right)=0.879\ {\rm m}\] 

(b) The gravitational potential energy of mass $m$ at height $h$ is $U_{grav}=mgh$, so
\[U_{grav}=25\times 9.8\times 0.879=215\ {\rm J}\] 

(c) Using the conservation of mechanical energy between the releasing point and the bottom of the swing, we get 
\[E_{top}=E_{bot}\] 
\[\Rightarrow mgh_i+\frac{1}{2}m\underbrace{v^2_i}_{v_i=0}=mg\underbrace{h_f}_{0}+\frac{1}{2}mv^2_f\] 
\[\Rightarrow v_f=\sqrt{2gh}=\sqrt{2\times 9.8\times 0.879}=4.15\frac{{\rm m}}{{\rm s}}\] 
 

(a) Since there is no friction in the system, so the mechanical energy of the cylinder is conserved so using this fact between initial and final points, we get \begin{gather*} E_i=E_f \\\\ mgh=\frac{1}{2}mv^2+\underbrace{\frac{1}{2}I\omega^2}_{\begin{array}{c}rotational \\ kinetic\ energy \end{array}}\end{gather*}
The required condition for rolling without slipping is $v=r\omega$ where $v$ is the tangential velocity of the cylinder (in this case the center of mass velocity) so
\begin{gather*}mgh=\frac{1}{2}mv^2+\frac{1}{2}\left(\frac{1}{2}mr^2\right){\left(\frac{v}{r}\right)}^2=\frac{3}{4}mv^2\\\\ \Rightarrow v_{CM}=\sqrt{\frac{4}{3}gh}\end{gather*}
Where we have used the moment of inertial of a hollow cylinder about the central axis i.e. $I_{cyl}=\frac{1}{2}mr^2$.
\[\Rightarrow v_{CM}=\sqrt{\frac{4}{3}(9.8)(1)}=3.61\, {\rm m/s}\] 
(b) Now relate the initial and final velocities of the cylinder with the constant acceleration of it by the following equation
\begin{align*}v^2-v^2_0&=2a_{CM}\Delta s\\\\ \Rightarrow \ a_{CM}&=\frac{v^2}{2\Delta s}\\\\&=\frac{v^2{\sin  20^\circ}}{2h}\\\\&=\frac{{\left(3.61\right)}^2{\sin 20^\circ}}{2\times 1}\\\\&=2.23\,{\rm \frac{m}{s^2}}\end{align*}
From the geometry we see that the distance traveled by the cylinder $\Delta s$ is $h/{\sin  20^\circ}$. 

(c) In part (b) the acceleration of the cylinder is obtained, so apply Newton's second law to the cylinder and find the desired quantity.
\begin{gather*} \Sigma F=ma\\\\ mg\,{\sin  20{}^\circ}-\mu_s\underbrace{mg\,{\cos 20^\circ}}_{N}=ma_{CM}\end{gather*} Solving the above for the unknown $\mu_s$, we have \begin{align*} \mu_s&=\frac{g\,{\sin  20^\circ}-a_{CM}}{g\,{\cos  20^\circ}}\\\\&=\frac{9.8\times {\sin 20^\circ}-2.23}{9.8\times {\cos  20^\circ}}\\\\&=0.121 \end{align*}
Important note: in the rolling without slipping motion, the ball is instantaneously at rest where it contacts the ground so we have used the static friction force equation. 
 

Solution 1 (kinematic language): use the kinematic relationships to find the speed of each rock after hitting the ground.

Let us consider the starting point as the reference point, thus the landing point has the coordinate ($x=?,y=-11\, {\rm m}$).

The first rock is thrown horizontally, so its motion is similar to a projectile with an angle of $\theta=0^\circ $. In the projectile motion, at every point, the velocity have two components as $v_x=v_{0x}=v_0\,{\cos  \theta}$ and $v_y=v_0\,{\sin \theta}-gt$. Thus first calculate the elapsed time to the hitting point as follows 
\begin{gather*} y=-\frac{1}{2}gt^2+v_0{\sin  \theta}\ t+y_0 \\\\  -11=-\frac{1}{2}(9.8)t^2+5t\,{\sin  0^\circ}+0 \\\\ \Rightarrow \quad t_{tot}=\sqrt{\frac{22}{9.8}}=1.5\, {\rm s}\end{gather*} 
Put it into the  $y$ component of the velocity $v_y=5\,{\sin  0^\circ}-9.8\times 1.5=-14.7\, {\rm m/s}$. The negative indicates that the rock is moving toward the ground. Using the Pythagorean theorem, we get the velocity of the horizontally thrown rock as \begin{align*} v&=\sqrt{v^2_x+v^2_y} \\\\ &=\sqrt{(5\cos 0^\circ)^2+(-14.7)^2} \\\\ &=15.52\,\rm m/s \end{align*} 
For the one which is thrown vertically, use the following kinematic relation to find its final velocity at the hitting point \begin{gather*} v^2-v^2_0=-2gh \\\\ v^2-5^2=-2(9.8)(-11) \\\\ \Rightarrow \quad v=15.52\, \rm m/s \end{gather*} As we can see, these two speeds are the same because they have the same mechanical energy at the moment of launch.

Solution 2 (energy approach):  
\[E_i=E_f\Rightarrow \frac{1}{2}mv^2_i+mgh_i=mg\underbrace{h_f}_{0}+\frac{1}{2}mv^2_f\] 
Dropping the mass of the rock and rearranging, we obtain 
\[v_f=\sqrt{v^2_i+2gh_i}=\sqrt{5^2+2(9.8)(11)}=15.52\frac{{\rm m}}{{\rm s}}\]