AP Physics

Unit 4 - Energy




You're a Phy Pro Member

Supercharge UBQ with

0 attempts

0% avg

UBQ Credits

Verfied Answer
Verfied Explanation 0 likes
Step Derivation/Formula Reasoning
1 KE_{\text{initial}} = \frac{1}{2} m v^2 Calculate the initial kinetic energy of the vehicle using its speed before it halts. Here v = 12.3 \text{ m/s}.
2 W_{\text{friction}} = f_k \cdot d = \mu_k \cdot m \cdot g \cdot \cos(\theta) \cdot d The work done by friction, where \mu_k = 0.650 is the coefficient of kinetic friction, g = 9.8 \text{ m/s}^2 is acceleration due to gravity, \theta = 18^\circ, and d is the distance the vehicle slides.
3 W_{\text{gravity}} = m \cdot g \cdot \sin(\theta) \cdot d The work done by gravity while the vehicle moves down the incline.
4 KE_{\text{initial}} = W_{\text{friction}} + W_{\text{gravity}} By the work-energy principle, the initial kinetic energy is converted into work done against friction plus the work done by gravity.
5 \frac{1}{2} m v^2 = \mu_k mg\cos(\theta)d + mg\sin(\theta)d Substitute expressions from steps 1, 2, and 3 into the work-energy equation.
6 d = \frac{\frac{1}{2} v^2}{\mu_k g \cos(\theta) + g \sin(\theta)} Solve for d, distance the vehicle slides. Notice that mass m cancels out.
7 d = \frac{\frac{1}{2} (12.3)^2}{0.650 \times 9.8 \times \cos(18^\circ) + 9.8 \times \sin(18^\circ)} Substitute numerical values for v, \mu_k, g, and \theta to find the value of d that represents the distance the vehicle slides until it stops.
8 d  = 8.32 \,\text{m} Calculated value.

To solve part b, look at the equation derived in step 6 of part a. Notice that the angle and distance traveled up the incline are inversely proportional. This means the greater the angle \theta the shorter the distance traveled.

Step Derivation/Formula Reasoning
1 d' = \frac{\frac{1}{2} v^2}{\mu_k g \cos(27^\circ) + g \sin(27^\circ)} Recalculate the distance with the increased angle of 1.5 \times 18^\circ = 27^\circ.
2 d' = \frac{d}{\cos(27^\circ) + \tan(27^\circ) \cdot \sin(27^\circ)} Using the previous formula of d, we express the new sliding distance d' in terms of the old distance d.
3 d' = 7.41 \, \text{meters} At an angle of 27^\circ the vehicale would slide up 7.41 meters, which is less than the orginal dsitance of 8.32 meters.
3 \frac{d'}{d} The ratio \frac{d'}{d} shows how much further the vehicle would slide relative to d.
4 \frac{7.41}{8.32} = .89 Thus the new distance d' = .89d

Part (c):

Step Derivation/Formula Reasoning
1 W_{\text{gravity,up}} = -m \cdot g \cdot \sin(\theta) \cdot d The work done by gravity as the vehicle slides up the incline is negative since gravity opposes the motion.
2 \frac{1}{2} m v^2 = \mu_k m g \cos(\theta) d – m g \sin(\theta) d Work-energy principle applied while moving up. The kinetic energy has to overcome both friction and an upward gravity force.
3 d_{\text{up}} = \frac{\frac{1}{2} v^2}{\mu_k g \cos(\theta) – g \sin(\theta)} Solve for the distance the vehicle would slide up the incline.
4 d_{\text{up}} < d The distance d_{\text{up}} will be lesser than d since gravity now acts against the motion, reducing the sliding distance relative to sliding down.

Need Help? Ask Phy To Explain This Problem

Phy can also check your working. Just snap a picture!

Simple Chat Box
NEW Smart Actions

Topics in this question

See how Others Did on this question | Coming Soon

Discussion Threads

Leave a Reply

  1. d = 8.32 \,\text{meters}
  2. d' = .89d

Nerd Notes

Discover the world's best Physics resources

Continue with

By continuing you (1) agree to our Terms of Sale and Terms of Use and (2) consent to sharing your IP and browser information used by this site’s security protocols as outlined in our Privacy Policy.

Sign In to View Your Questions

Share This Question

Enjoying UBQ? Share the 🔗 with friends!

Link Copied!
Made By Nerd-Notes.com
\Delta x = v_i t + \frac{1}{2} at^2F = ma
v = v_i + atF_g = \frac{G m_1m_2}{r^2}
a = \frac{\Delta v}{\Delta t}f = \mu N
R = \frac{v_i^2 \sin(2\theta)}{g} 
Circular MotionEnergy
F_c = \frac{mv^2}{r}KE = \frac{1}{2} mv^2
a_c = \frac{v^2}{r}PE = mgh
 KE_i + PE_i = KE_f + PE_f
MomentumTorque and Rotations
p = m v\tau = r \cdot F \cdot \sin(\theta)
J = \Delta pI = \sum mr^2
p_i = p_fL = I \cdot \omega
Simple Harmonic Motion
F = -k x
T = 2\pi \sqrt{\frac{l}{g}}
T = 2\pi \sqrt{\frac{m}{k}}
gAcceleration due to gravity, typically 9.8 , \text{m/s}^2 on Earth’s surface
GUniversal Gravitational Constant, 6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2
\mu_k and \mu_sCoefficients of kinetic (\mu_k) and static (\mu_s) friction, dimensionless. Static friction (\mu_s) is usually greater than kinetic friction (\mu_k) as it resists the start of motion.
kSpring constant, in \text{N/m}
M_E = 5.972 \times 10^{24} , \text{kg} Mass of the Earth
M_M = 7.348 \times 10^{22} , \text{kg} Mass of the Moon
M_M = 1.989 \times 10^{30} , \text{kg} Mass of the Sun
VariableSI Unit
s (Displacement)\text{meters (m)}
v (Velocity)\text{meters per second (m/s)}
a (Acceleration)\text{meters per second squared (m/s}^2\text{)}
t (Time)\text{seconds (s)}
m (Mass)\text{kilograms (kg)}
VariableDerived SI Unit
F (Force)\text{newtons (N)}
E, PE, KE (Energy, Potential Energy, Kinetic Energy)\text{joules (J)}
P (Power)\text{watts (W)}
p (Momentum)\text{kilogram meters per second (kgm/s)}
\omega (Angular Velocity)\text{radians per second (rad/s)}
\tau (Torque)\text{newton meters (Nm)}
I (Moment of Inertia)\text{kilogram meter squared (kgm}^2\text{)}
f (Frequency)\text{hertz (Hz)}

General Metric Conversion Chart

Example of using unit analysis: Convert 5 kilometers to millimeters. 

  1. Start with the given measurement: \text{5 km}

  2. Use the conversion factors for kilometers to meters and meters to millimeters: \text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}

  3. Perform the multiplication: \text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}} = 5 \times 10^3 \times 10^3 \, \text{mm}

  4. Simplify to get the final answer: \boxed{5 \times 10^6 \, \text{mm}}



Power of Ten




















(Base unit)


Deca- or Deka-


















  1. Some answers may be slightly off by 1% depending on rounding, etc.
  2. Answers will use different values of gravity. Some answers use 9.81 m/s2, and other 10 m/s2 for calculations.
  3. Variables are sometimes written differently from class to class. For example, sometime initial velocity v_i is written as u ; sometimes \Delta x is written as s .
  4. Bookmark questions that you can’t solve so you can come back to them later. 
  5. Always get help if you can’t figure out a problem. The sooner you can get it cleared up the better chances of you not getting it wrong on a test!

Phy Pro

The most advanced version of Phy. Currently 50% off, for early supporters.


per month

Billed Monthly. Cancel Anytime.

Trial  –>  Phy Pro

Error Report

Sign in before submitting feedback.

You can close this ad in 5 seconds.

Ads show frequently. Upgrade to Phy Pro to remove ads.

You can close this ad in 7 seconds.

Ads display every few minutes. Upgrade to Phy Pro to remove ads.

You can close this ad in 5 seconds.

Ads show frequently. Upgrade to Phy Pro to remove ads.

Jason here! Feeling uneasy about your next physics test? We will help boost your grade in just two hours.

We use site cookies to improve your experience. By continuing to browse on this website, you accept the use of cookies as outlined in our privacy policy.