# AP Physics C Mechanics: Unit 3 Practice Test — Work, Energy, & Power

AP Physics C: Mechanics practice test 3.

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 Question 1

### A boy pedals a cycle to reach a speed of $6 \ m/s$. The force needed to keep the cycle moving at the same speed is $40 \ N$. What is the power delivered by the boy to keep the cycle moving forward?

 A $460 \ W$ B $330 \ W$ C $196 \ W$ D $400 \ W$ E $240 \ W$
Question 1 Explanation:
Power delivered $=Fv \ \cos⁡θ$

Using $v=6 \ m/s, F=40 \ N$ and $θ=0°$

Power delivered $=40×6×1$

$=240 \ W$
 Question 2

### What is the total work done by the force on the object?

 A $32 \ J$ B $64 \ J$ C $50 \ J$ D $46 \ J$ E $38 \ J$
Question 2 Explanation:
Work done = Area under the curve $=A_I+A_{II}$

$A_I=\dfrac{1}{2}×(2+5)×4=14 \ J$

$A_{II}=\dfrac{1}{2}×(1+5)×6=18 \ J$

Total work done $=14+18=32 \ J$
 Question 3

### A particle of mass $0.5 \ kg$ is at rest and is constrained to move parallel to the $y-axis$. If a force of $(10\skew{2.5}\hat{i} + 6\skew{4}\hat{j}) \ N$ starts acting on the object from $t=0 \ s$, what is the work done on the object from $t=0 \ s$ to $t=4 \ s$?

 A $724 \ J$ B $960 \ J$ C $576 \ J$ D $441 \ J$ E $660 \ J$
Question 3 Explanation:
Force acting in the $y$ direction $=6 \ N$

Acceleration of the object in the y direction $=\dfrac{6}{0.5}=12 \ m/s^2$

Displacement in the $y$ direction $=h$

And, $h=ut+\dfrac{1}{2} at^2$

Using $u=0 \ m/s, t=4 \ s$ and $a=12 \ m/s^2$

$h=0+\dfrac{1}{2}×12×4^2=96 \ m$

Work done $=F×h$

$=6×96$

$=576 \ J$

 Question 4

### A block at rest is pulled by a force acting at an angle of $60°$ to the horizontal. The block displaces by $23 \ m$ horizontally without being accelerated by the force. If the work done by the force is $230 \ J$, what is the force acting on the block?

 A $(17.3 \skew{2.3}\hat{i} + 20 \skew{4.5}\hat{j} ) \ N$ B $(10 \skew{2.3}\hat{i} + 17.3 \skew{4.5}\hat{j} ) \ N$ C $(20 \skew{2.3}\hat{i} +10 \skew{4.5}\hat{j} ) \ N$ D $(17.3 \skew{2.3}\hat{i} + 10 \skew{4.5}\hat{j} ) \ N$ E $(10 \skew{2.3}\hat{i} +20 \skew{4.5}\hat{j} ) \ N$
Question 4 Explanation:
The free body diagram for the object is:

Work done on the object $W=F \ \cos⁡θ×s$

Using $W=230 \ J, s=23 \ m$ and $θ=60°$

$230=F \ \cos⁡ 60°×23$

$F=20 \ N$

Thus, $F_x=20 \ \cos⁡60°=10 \ N$ and $F_y=20 \ \sin⁡60°=17.3 \ N$

$\overrightarrow{F}=(10\skew{2}\hat{i} + 17.3 \skew{5}\hat{j} ) N$
 Question 5

### The block’s speed changes from $15 \ m/s$ to $3 \ m/s$ in $5 \ s$. If the weight of the block is equal to $100 \ N$, what is the coefficient of kinetic friction between the block and the table?

 A $0.17$ B $0.14$ C $0.19$ D $0.24$ E $0.22$
Question 5 Explanation:
Deceleration $a=\dfrac{3 - 15}{5}=-2.4 \ m/s^2$

Mass of the block $=\dfrac{100}{10}=10 \ kg$

Frictional force on the block $=ma=-24 \ N$

($-ive$ sign shows that the force is in the opposite direction that of the motion)

Using $f_r=24 \ N, N=W=100 \ N$ in $f_r=μ_k \ N$ gives,

$μ_k=\dfrac{24}{100}=0.24$
 Question 6

### What is the change in the total kinetic energy of the two-block system when the left block reaches the pulley after being allowed to move from the above configuration?(Here, $g$ is the acceleration due to gravity)

 A $2.4 \ g$ B $3 \ g$ C $1.2 \ g$ D $2 \ g$ E $2.8 \ g$
Question 6 Explanation:
Initial kinetic energy of the two-block system $=0$

Initial potential energy of the two-block system $=2×g×2+3×g×2=10 \ g$

New configuration of the two-block system is:

New potential energy of the two-block system $=2×g×4+3×g×0=8 \ g$

New kinetic energy of the two-block system

$=(10g+0)-8g=2 \ g$

Change in the kinetic energy of the two-block system $=2g-0=2 \ g$
 Question 7

### What is the change in the kinetic energy of the system when the object moves from $x=2 \ m$ to $x=6 \ m$?

 A $24 \ J$ B $30 \ J$ C $44 \ J$ D $36 \ J$ E $18 \ J$
Question 7 Explanation:
Work done = Change in the kinetic energy of the system

Work done $=\int_{x=2}^{x=6} \ f(x)dx$

$=\int_{x=2}^{x=6} \ (2x+1)dx$

$= x^2+x |_2^6$

$=(6^2-2^2 )+(6-2)$

$=32+4$

$=36 \ J$
 Question 8

### The track has a single circular loop of diameter $3 \ m$ after the curve. What is the minimum height from which the toy should be left so that it completes the circle without losing contact at any point on it?

 A $3.75 \ m$ B $3.00 \ m$ C $4.55 \ m$ D $3.45 \ m$ E $3.95 \ m$
Question 8 Explanation:
The minimum velocity (at the topmost point of the circular track) at which the toy can complete the track without falling down is:

$v=\sqrt{gr} \$ thus $\ v^2=1.5g$

From the conservation of energy,

$PE_{topmost}=PE_{top}+KE_{top}$

$0.2×g×h=0.2×g×3+0.5×0.2×1.5×g$

$0.2h=0.75→h=3.75 \ m$
 Question 9

### What is the tension force acting on the mass when it is at point $A$?

 A $9.89 \ N$ B $10.28 \ N$ C $7.99 \ N$ D $13.33 \ N$ E $11.21 \ N$
Question 9 Explanation:
Total energy of the pendulum at point $C=-0.5×10×2 \ \cos⁡ 60°=-5 \ J$

Total energy of the pendulum at point $A=\dfrac{1}{2} \ mv^2-mgr \ \cos⁡ 30°$

Since, total mechanical energy is a constant

$\dfrac{1}{2} \ mv^2=-5+0.5×10×2×\cos⁡ 30°=3.66 \ J$

$v^2=14.64$

At point $A$,

$T=mg \ \cos⁡ 30° +\dfrac{mv^2}{r}$

$T=0.5×10×\dfrac{\sqrt{3}}{2}+\dfrac{0.5×14.64}{2}$

$T=7.99 \ N$
 Question 10

### If the mass of the block is equal to $2 \ kg$, then what is the energy loss due to frictional force as the block slides down from the top to the bottom?

 A $33.01 \ A$ B $21.08 \ A$ C $22.52 \ A$ D $28.83 \ A$ E $19.02 \ A$
Question 10 Explanation:
Energy loss = Work done by the frictional force

Work done $=\int_0^{10} \ f_r (x) \ dx$

$=\int_0^{10} \ A \sqrt{x} \ dx$

$=A \int_0^{10} \ \sqrt{x} \ dx$

$=\dfrac{2A}{3}×10^{3/2}$

$=21.08 \ A$
 Question 11

### If the velocity of the block at the bottom of the inclined plane is $\sqrt{21} \ m/s$, then what is the value $A$?

 A $2.95$ B $1.90$ C $4.20$ D $3.33$ E $3.75$
Question 11 Explanation:
Total energy at the top of the inclined plane
$=2×10×5=100 \ J$

Total energy at the bottom of the inclined plane
$=\dfrac{1}{2}×2×(\sqrt{21})^2=21 \ J$

Energy loss $=100-21=79 \ J$
$21.08A=79$
$→A=3.75$
 Question 12

### 10% of the total mechanical energy of the ball is lost by the time it reaches the topmost point of its trajectory. The height h that the ball rises is equal to,(Assume the spring constant as $\dfrac{2mg}{l}$ and the energy transferred to the ball is 100%)

 A $\dfrac{0.9x^2}{l}-0.5 \ (l-x)$ B $\dfrac{0.9x^2}{l}+0.9 \ (x-l)$ C $\dfrac{1.2x^2}{l}+1.2 (l-x)$ D $\dfrac{0.9x^2}{l}+0.9 \ (l-x)$ E $\dfrac{0.5x^2}{l}+0.5 \ (l-x)$
Question 12 Explanation:
Initial energy of the ball $=\dfrac{1}{2} \ kx^2+mg \ (l-x)$

Final energy of the ball $=mgh$

Final energy of the ball $=90 \%$ of the initial energy

$mgh=0.9 \left(\dfrac{1}{2} kx^2+mg(l-x)\right)$

$h=0.9\left(\dfrac{1}{2} \ \dfrac{k}{mg} \ x^2+(l-x)\right)$

$h=\dfrac{0.9x^2}{l}+0.9(l-x)$
 Question 13

### The potential energy of a 1D system is equal to $U(x)=-x \left(1+ \dfrac{2}{x} e^{-x} \right)$. Which of the below graphs shows the force $f$ as a function of the position x?

 A B C D E
Question 13 Explanation:
$U(x)=-x \left(1+\dfrac{2}{x} e^{-x} \right)$

Simplifying the potential gives,

$U(x)=-x-2e^{-x}$

$f(x)=-\dfrac{dU}{dx}$

Thus, $f(x)=-\dfrac{d}{dx} (-x-2e^{-x} )$

$f(x)=-(-1+2e^{-x} )$

$f(x)=1-2e^{-x}$

Since $f(0)=-1$ and as $x→+∞, f(x)→1$

Which is true for Option E
 Question 14

### (Assume the total mechanical energy of the system is a constant)

 A $5 \ m/s$ B $4 \ m/s$ C $6 \ m/s$ D $8 \ m/s$ E $7 \ m/s$
Question 14 Explanation:
Total mechanical energy of the system $=10 \ J$

Total mechanical energy of the system at point
$X=-5+KE$

Thus, $KE=10+5=15 \ J$

$v=\sqrt{\dfrac{2×KE}{m}}$

$=\sqrt{\dfrac{2×15}{1.2}}$

$=\sqrt{25}$

$=5 \ m/s$
 Question 15

### What is the power delivered to the block as it goes from time $t=2 \ s$ to $t=5 \ s$?(Hint: Use $P=∫_{t_1}^{t_2} F(t)a(t)dt)$

 A $4.50 \ W$ B $3.67 \ W$ C $3.33 \ W$ D $5.25 \ W$ E $6.20 \ W$
Question 15 Explanation:
Acceleration of the block $a(t)=\dfrac{F(t)}{m}=\dfrac{5}{t + 1}$

$P=\int_{t_1}^{t_2} F(t)a(t) \ dt$

$=\int_2^5 \dfrac{4}{t+1} \ \dfrac{5}{t+1} \ dt$

$=20\int_2^5 \dfrac{1}{(t+1)^2} \ dt$

$=20 \dfrac{(t+1)^{-1}}{-1} |_2^5$

$=-20\left( \dfrac{1}{6} - \dfrac{1}{3}\right)$

$=3.33 \ W$
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