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AP Physics 1 : Work, Energy, and Power

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Example Questions

Example Question #81 : Work, Energy, And Power

What is the mass of a bowling ball that maintains an average kinetic energy of 75J while traveling at $7\frac{m}{s}$ ?

Possible Answers:

3.1kg

2.1kg

4.1kg

1.1kg

Correct answer:

3.1kg

Explanation:

For this question we will use the kinetic energy equation.

$K_{e}=\frac{1}{2}m(v^{2})$

Plug in and solve for the mass of the bowling ball.

$75j=\frac{1}{2}m(7\frac{m}{s})^{2}$

m=3.06Kg

The mass of the bowling ball is 3.1Kg .

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Example Question #81 : Work, Energy, And Power

A child weighing in an 30kg is walking at $3.0\frac{m}{s}$ . What is his or her average kinetic energy?

Possible Answers:

100J

200J

135J

50J

Correct answer:

135J

Explanation:

For this question use the equation for kinetic energy.

$K_{e}=\frac{1}{2}m(v^{2})$

Plug in and solve for the the kinetic energy.

$K_{e}=\frac{1}{2}(30Kg)3.0\frac{m}{s}$

$K_{e}=135j$

The child will have a kinetic energy equal to 135j .

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Example Question #33 : Kinetic Energy

A box of mass 50kg is initially at rest, and is then subject to 200N of a directing horizontal pulling force. Neglecting friction and air resistance, what is the kinetic energy of the box after 5 s ?

Possible Answers:

10kJ

5kJ

1kJ

100kJ

Correct answer:

10kJ

Explanation:

First, solve for the acceleration of the box using Newton's second law.

f=ma

Plug in and solve for the acceleration.

200N=50kg(a)

$a=4\frac{m}{s^{2}}$

Now use a kinematic equation to solve for the final velocity.

$v_{f}=v_{o}+a\Delta t$

Plug in and solve for the final velocity.

$v_{f}=0\frac{m}{s}+4\frac{m}{s^{2}}(5s)$ $v_{f}=20\frac{m}{s}$

Finally, we will use the kinetic energy equation to determine the kinetic energy of the box after 5 seconds .

$K_{e}=\frac{1}{2}m(v^{2})$

Plug in and solve for the kinetic energy.

$K_{e}=\frac{1}{2}50Kg(20\frac{m}{s})^{2}$

$K_{e}=10,000j$

The final kinetic energy of the box will be 10Kj .

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Example Question #34 : Kinetic Energy

Two space ships are racing deep in space. At t=0 , Spaceship A is at the origin while traveling at $<600,400> \frac{m}{s}$ . It is coasting, with it’s rockets off. Spaceship B is at <-600,-400>m . It’s traveling at $<150,100>\frac{m}{s}$ . It’s rockets are firing with a constant force of <1.5*10^6N, 1*10^6N> . Both rockets have a mass of 1.5*10^5 kg .

What is the total kinetic energy of the two spaceships at t=0 ?

Possible Answers:

$KE_{total}=3.95*10^{10}J$

None of these

$KE_{total}=8.44*10^{10}J$

$KE_{total}=6.05*10^{10}J$

$KE_{total}=4.15*10^{10}J$

Correct answer:

$KE_{total}=4.15*10^{10}J$

Explanation:

Finding net velocity of Spaceship :

$v_{net}=\sqrt{v_x^2+v_y^2}$

$v_{net}=\sqrt{600^2+400^2}$

$v_{net}=721\frac{m}{s}$

Finding net velocity of Spaceship :

$v_{net}=\sqrt{v_x^2+v_y^2}$

$v_{net}=\sqrt{150^2+100^2}$

$v_{net}=180\frac{m}{s}$

$KE_{total}=KE_1+KE_2$

$KE_{total}=.5m_Av_A^2+.5m_Bv_B^2$

$KE_{total}=.5*1.5*10^5*721^2+.5*1.5*10^5*180^2$

$KE_{total}=4.15*10^{10}Joules$

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Example Question #35 : Kinetic Energy

Moment of inertia of disk: .5MR^2

A car with wheels of mass 25kg and wheels of radius .25m is traveling at $50\frac{m}{s}$ . Treating the wheels as disks of uniform mass density, calculate the rotational kinetic energy of one wheel.

Possible Answers:

$KE_{rot}=150J$

$KE_{rot}=15625J$

None of these

$KE_{rot}=35789J$

$KE_{rot}=1338J$

Correct answer:

$KE_{rot}=15625J$

Explanation:

Finding angular velocity:

$50\frac{m}{s}*\frac{1 rotation}{2*\pi*.25m}*\frac{2\pi radians}{rotation}=200\frac{radians}{second}$

Using

I=.5MR^2

and

$KE_{rot}=.5I\omega^2$

Combining equations and plugging in values:

$KE_{rot}=.5*.5*25*.25^2*200^2$

$KE_{rot}=15625Joules$

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Example Question #82 : Work, Energy, And Power

A soccer ball at rest of mass 4kg is accelerated at $3\frac{m}{s}^2$ for . What is the final kinetic energy in Joules of the ball after ?

Possible Answers:

KE=1600J

KE=1550J

None of these

KE=1322J

KE=1458J

Correct answer:

KE=1458J

Explanation:

First find the final speed of the ball using $V=V_{0}+at$ . Then, use KE=.5mv^2 to find the kinetic energy after 9s.

$V=(3\frac{m}{s^2})(9s)=27\frac{m}{s}$

$KE=.5(4kg)(27\frac{m}{s})^2=\boldsymbol{1458J}$

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Example Question #37 : Kinetic Energy

A 1200 kg missile has a net propulsion force of 60 kN . Over a brief period, the missile speeds up uniformly from an initial velocity $30\frac{m}{s}$ to $40\frac{m}{s}$ . Assume that the missile will not lose mass during fuel consumption and so the net propulsion remains constant. What is the net work done on the rocket in kilojoules?

Possible Answers:

420kJ

333kJ

300kJ

275kJ

268kJ

Correct answer:

420kJ

Explanation:

Find the KE of the missile at each velocity. The difference or change in KE is the net work done on the missile.

KE=.5(1200kg)30^2=540,000J

KE=.5(1200kg)40^2=960,000J

$\DeltaKE=960,000-540,000=420,000J$

Divide by 1000 to convert to . 420kJ

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Example Question #81 : Work, Energy, And Power

A bullet with mass .05 kg hits the bullseye at $300\frac{m}{s}$ . Assuming no loss of energy due to heat or frictional forces, how much energy in Joules did the bullet transfer to the target?

Possible Answers:

2250J

2900J

2400J

2150J

2500J

Correct answer:

2250J

Explanation:

Find the kinetic energy:

$KE=.5mv^2\rightarrow .5(0.05kg)(300\frac{m}{s})^2=\boldsymbol{2250J}$

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Example Question #39 : Kinetic Energy

A sledder of mass m = 60kg is traveling down a hill with slope of $40^{\circ}$ above the horizontal. If the sledder starts from rest and has an average frictional force of 150N , what is the velocity of the sledder after traveling 70m along the slope?

$g=10\frac{m}{s^2}$

Possible Answers:

$23.5 \frac{m}{s}$

$15.6\frac{m}{s}$

$31.2\frac{m}{s}$

$26.0\frac{m}{s}$

$19.7\frac{m}{s}$

Correct answer:

$23.5 \frac{m}{s}$

Explanation:

We can begin with the expression for conservation of energy for this problem:

E=U_i+K_i+W=U_f+K_f

We are told that the sledder begins from rest, so we can eliminate initial kinetic energy. Also, if we say that the final height of the sledder is 0, we can eliminate final potential energy. Also, the only work being done is from friction, and the work is being removed from the system, so it will be negative. Therefore, we get:

U_i+W_F=K_f

Substituting in expressions for each of these, we get:

$mgh_i-F_Fd=\frac{1}{2}mv_f^2$

Rearranging for final velocity:

$v_f = \sqrt{\frac{2(mgh_i-F_Fd)}{m}}$

Now we just need an expression for initial height:

$h_i = sin(\theta)d$

Plugging this in, we get:

$v_f = \sqrt{\frac{2(mgsin(\theta)d-F_Fd)}{m}}$

We have all of these values, so time to plug and chug:

$v_f = 23.5\frac{m}{s}$

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Example Question #40 : Kinetic Energy

Rollercoaster 2

The following train car has just went around the loop of this roller coaster and is now travelling to the left. The train car weighs $545\ kg$ and is travelling at a speed of $52.0\ \frac{m}{s}$ .

What is the approximate kinetic energy of the train car?

Possible Answers:

$524\ kJ$

$596\ kJ$

$737\ kJ$

$912\ kJ$

$651\ kJ$

Correct answer:

$737\ kJ$

Explanation:

Kinetic energy is defined by the following equation:

$KE=\frac{1}{2}mv^{2}$

where is mass, and is velocity.

In this question, you are given the mass and velocity, so by plugging in the given values it is possible to find the kinetic energy (in Joules).

$KE=\frac{1}{2}(545)(52.0^{2})$

$KE\approx 737\ kJ$

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AP Physics 1 : Work, Energy, and Power

Study concepts, example questions & explanations for AP Physics 1

All AP Physics 1 Resources

Example Questions

Example Question #81 : Work, Energy, And Power

Example Question #81 : Work, Energy, And Power

Example Question #33 : Kinetic Energy

Example Question #34 : Kinetic Energy

Example Question #35 : Kinetic Energy

Example Question #82 : Work, Energy, And Power

Example Question #37 : Kinetic Energy

Example Question #81 : Work, Energy, And Power

Example Question #39 : Kinetic Energy

Example Question #40 : Kinetic Energy

All AP Physics 1 Resources

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