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

# Part (a) Free Body Diagram Explanation

Please visualize or draw out the FBD as described:

Gravity ( \vec{mg} ): Acts directly downwards. With a mass of m = 90 + 12 = 102 kg.
Normal Force ( \vec{N} ): Acts perpendicular to the surface of the incline.
Frictional Force ( \vec{f} ): Acts parallel to the incline against the direction of motion during ascent and in the opposite direction of velocity during descent.
Component of Gravitational Force down the incline ( \vec{mg} \sin(\theta) ): Helps in descending and resists during ascending. Here \theta = 30^\circ .

# Part (b) Calculation of Work Done by Friction to Stop the Bicycle

Step Derivation/Formula Reasoning
1 F_f = \mu_k N Frictional force, F_f , is the product of the coefficient of kinetic friction, \mu_k , and the normal force, N .
2 N = mg \cos(\theta) Normal force is the component of the gravitational force perpendicular to the incline.
3 F_f = 0.7 \times 102 \times 9.8 \times \cos(30^\circ) Calculate F_f using \mu_k = 0.7 , m = 102 kg, g = 9.8 \, \text{m/s}^2 , and \theta = 30^\circ .
4 W_f = -F_f d Work done by friction, W_f , is the product of the frictional force and the distance, d , over which it acts, with a negative sign indicating work done against the motion.
5 W_f = -0.7 \times 102 \times 9.8 \times \cos(30^\circ) \times 9 Substitute values to calculate the work done. The distance d = 9 m.
6 W_f \approx -5454 \, \text{J} Calculated work done by friction; it’s negative as it opposes the direction of motion.

# Part (c) Explanation – Difficulty of Traveling Up vs. Down the Incline

Step Derivation/Formula Reasoning
1 F_{\text{gravity, down}} = mg \sin(\theta) Component of gravitational force along the incline that assists in descending and opposes during ascending.
2 F_{\text{friction, up}} = \mu_s N Static friction opposing the upward motion, which is higher due to increased force requirements.
3 F_{\text{net, up}} = F_{\text{gravity, down}} + F_{\text{friction, up}} Sum of forces opposing the ascent, both the gravitational pull back down the incline and the frictional force.
4 F_{\text{net, down}} = F_{\text{gravity, down}} – F_{\text{friction, down}} Net force during descent is reduced because friction (now kinetic) is less than static friction and gravity assists in motion.
5 Comparison The cyclist finds it harder to travel up due to higher net opposing force (more friction and gravity opposing motion).

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. FBD with 3 forces.
  2. W_f \approx -5454 \, \text{J}
  3. The cyclist finds it harder to travel up due to higher net opposing force (more friction and gravity opposing motion).

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.