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?

$460 \ W$ | |

$330 \ W$ | |

$196 \ W$ | |

$400 \ W$ | |

$240 \ W$ |

Question 1 Explanation:

Power delivered $=Fv \ \cosθ$

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

Power delivered $=40×6×1$

$=240 \ W$

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

Power delivered $=40×6×1$

$=240 \ W$

Question 2 |

### A position-dependent force (in 1D) acts on an object of mass $m$. The graph shows the variation in the force as the object displaces from its initial position.

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

$32 \ J$ | |

$64 \ J$ | |

$50 \ J$ | |

$46 \ J$ | |

$38 \ J$ |

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 $=14+18=32 \ J$

$A_I=\dfrac{1}{2}×(2+5)×4=14 \ J$

$A_{II}=\dfrac{1}{2}×(1+5)×6=18 \ J$

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 $t=4 \ s$?

$724 \ J$ | |

$960 \ J$ | |

$576 \ J$ | |

$441 \ J$ | |

$660 \ J$ |

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 $=F×h$

$=6×96$

$=576 \ J$

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?

$(17.3 \skew{2.3}\hat{i} + 20 \skew{4.5}\hat{j} ) \ N$ | |

$(10 \skew{2.3}\hat{i} + 17.3 \skew{4.5}\hat{j} ) \ N$ | |

$(20 \skew{2.3}\hat{i} +10 \skew{4.5}\hat{j} ) \ N$ | |

$(17.3 \skew{2.3}\hat{i} + 10 \skew{4.5}\hat{j} ) \ N$ | |

$(10 \skew{2.3}\hat{i} +20 \skew{4.5}\hat{j} ) \ N$ |

Question 4 Explanation:

The free body diagram for the object is:

Work done on the object $W=F \ \cosθ×s$

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

$230=F \ \cos 60°×23$

$F=20 \ N$

Thus, $ F_x=20 \ \cos60°=10 \ N$ and $F_y=20 \ \sin60°=17.3 \ N$

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

Work done on the object $W=F \ \cosθ×s$

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

$230=F \ \cos 60°×23$

$F=20 \ N$

Thus, $ F_x=20 \ \cos60°=10 \ N$ and $F_y=20 \ \sin60°=17.3 \ N$

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

Question 5 |

### A heavy block moving on a table decelerates due to the frictional force.

### 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?

$0.17$ | |

$0.14$ | |

$0.19$ | |

$0.24$ | |

$0.22$ |

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 $

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 |

### Two blocks are resting on a frictionless inclined plane, as shown below.

### 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)

$2.4 \ g$ | |

$3 \ g$ | |

$1.2 \ g$ | |

$2 \ g$ | |

$2.8 \ g$ |

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 $=2g-0=2 \ g$

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 |

### The net force acting on an object is given by the equation,

### $f(x)=(2x+1) \ N$

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

$24 \ J$
| |

$30 \ J$
| |

$44 \ J$
| |

$36 \ J$
| |

$18 \ J$ |

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) $

$=32+4$

$=36 \ J$

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 |

### A toy of mass $0.2 \ kg$ is left from the top of a pre-determined track of height $h m$.

### 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?

$3.75 \ m$ | |

$3.00 \ m$ | |

$4.55 \ m$ | |

$3.45 \ m$ | |

$3.95 \ m$ |

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} $

$0.2×g×h=0.2×g×3+0.5×0.2×1.5×g$

$0.2h=0.75→h=3.75 \ m$

$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 |

### George sets a pendulum in motion from point $C$. The length of the pendulum is equal to $2 \ m$, and the mass attached to the pendulum is $0.5 \ kg$.

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

$9.89 \ N$ | |

$10.28 \ N$ | |

$7.99 \ N$ | |

$13.33 \ N$ | |

$11.21 \ N$ |

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 $

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 |

**Questions 10 and 11 are based on the below information:**

### A weird positional dependent frictional force acts on a block sliding down from the top of an inclined plane of slanting length $10 \ m$ and height $5 \ m$. The dependence of the frictional force on the position is given as $f_r (x)=A \sqrt{x} \ N$ where $A$ is $a$ constant and $x$ is the distance from the top of the inclined plane.

### 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?

$33.01 \ A$ | |

$21.08 \ A$ | |

$22.52 \ A$ | |

$28.83 \ A$ | |

$19.02 \ A$ |

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 $

$=\dfrac{2A}{3}×10^{3/2}$

$=21.08 \ A$

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 |

### A weird positional dependent frictional force acts on a block sliding down from the top of an inclined plane of slanting length $10 \ m$ and height $5 \ m$. The dependence of the frictional force on the position is given as $f_r (x)=A \sqrt{x} \ N$ where $A$ is $a$ constant and $x$ is the distance from the top of the inclined plane.

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

$2.95$ | |

$1.90$ | |

$4.20$ | |

$3.33$ | |

$3.75$ |

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 $=100-21=79 \ J$

$21.08A=79$

$→A=3.75$

$=2×10×5=100 \ J$

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 |

### A ball of mass m is used to push down a spring of length $l$ by $x$. The system is then allowed to move from this configuration.

### 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%)

$\dfrac{0.9x^2}{l}-0.5 \ (l-x)$ | |

$\dfrac{0.9x^2}{l}+0.9 \ (x-l)$ | |

$\dfrac{1.2x^2}{l}+1.2 (l-x)$ | |

$\dfrac{0.9x^2}{l}+0.9 \ (l-x)$ | |

$\dfrac{0.5x^2}{l}+0.5 \ (l-x)$ |

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) $

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?

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

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 |

### A 1D system has a potential energy function, as shown below.

### A particle of mass $1.2 \ kg$ is released with zero initial velocity and with an energy of $10 \ J$. Using the graph, determine the velocity of the particle at point $X$.

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

$5 \ m/s$ | |

$4 \ m/s$ | |

$6 \ m/s$ | |

$8 \ m/s$ | |

$7 \ m/s$ |

Question 14 Explanation:

Total mechanical energy of the system $=10 \ J$

Total mechanical energy of the system at point

$X=-5+KE$

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

$v=\sqrt{\dfrac{2×KE}{m}}$

$=\sqrt{\dfrac{2×15}{1.2}}$

$=\sqrt{25}$

$=5 \ m/s$

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 |

### A non-constant, non-conservative force $F(t)$ acts on a block, as shown below.

### Here, $F(t)=\dfrac{4}{t + 1} \ N$ and $t≥0$.

### What is the power delivered to the block as it goes from time $t=2 \ s$ to $t=5 \ s$?

(Hint: Use $P=∫_{t_1}^{t_2} F(t)a(t)dt)$

$4.50 \ W$ | |

$3.67 \ W$ | |

$3.33 \ W$ | |

$5.25 \ W$ | |

$6.20 \ W$ |

Question 15 Explanation:

Acceleration of the block $a(t)=\dfrac{F(t)}{m}=\dfrac{5}{t + 1}$

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

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

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

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

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

$=3.33 \ W$

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

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

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

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

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

$=3.33 \ W$

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