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Unit 2 - Linear Forces

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First find the pushing force, which equal to the maximum force of static friction.

Step Formula Derivation Reasoning
1 [katex]F_{\text{static max}} = \mu_{s} F_{\text{normal}}[/katex] Maximum static friction force, where [katex]\mu_{s}[/katex] is the coefficient of static friction.
2 [katex]F_{\text{normal}} = mg[/katex] Normal force equals weight for horizontal motion.
3 [katex]F_{\text{static max}} = \mu_{s} mg[/katex] Substituting [katex]F_{\text{normal}}[/katex].

Now find the net force during the first 5 seconds. Use this to find the acceleration of the block in the first 5 seconds.

Step Formula Derivation Reasoning
1 [katex]F_{\text{net}} = F_{\text{push}} – F_{\text{kinetic}}[/katex] Net force is the difference between pushing force and kinetic friction.
2 [katex]F_{\text{push}} = \mu_{s} F_{\text{normal}}[/katex] Pushing force equal to maximum static friction.
3 [katex]F_{\text{kinetic}} = \mu_{k} F_{\text{normal}}[/katex] Kinetic friction force.
4 [katex]F_{\text{net}} = \mu_{s} mg – \mu_{k} mg[/katex] Substituting values for [katex]F_{\text{push}}[/katex] and [katex]F_{\text{kinetic}}[/katex].
5 [katex]F_{\text{net}} = \mu_{s} mg – \mu_{k} mg = ma [/katex] Set net force equation equal to ma. Then solve for a (acceleration)
6 a = 1 m/s2 Plug in values and solve

Final Speed at End of 5 Seconds

Step Formula Derivation Reasoning
1 [katex]v = u + \frac{F_{\text{net}}}{m}t[/katex] Kinematic equation for velocity. Note F/m is the acceleration (1 m/s2 ) from above.
2 [katex]v = 0 + \frac{(\mu_{s} – \mu_{k}) mg}{m} \times 5, \text{s}[/katex] Initial velocity [katex]u = 0[/katex]; substituting [katex]F_{\text{net}}[/katex] and [katex]t[/katex].
3 [katex]v = 5 m/s [/katex] Plug in values and find the velocity at the end of the 5 second push.

Time to Stop Due to Just Kinetic Friction (after you stop pushing)

Step Formula Derivation Reasoning
1 [katex]0 = v – \mu_{k} gt_{\text{stop}}[/katex] Final velocity [katex]0[/katex] when stopped, [katex]v[/katex] is final speed after push (5 m/s as found above).
2 [katex]t_{\text{stop}} = \frac{v}{\mu_{k} g}[/katex] Solving for time to stop [katex]t_{\text{stop}}[/katex].

Total Time

Step Formula Derivation Reasoning
1 [katex]t_{\text{total}} = t_{\text{push}} + t_{\text{stop}}[/katex] Sum of time with force applied and time to stop.

Let’s calculate [katex]t_{\text{total}}[/katex]. The calculations yield the following results:

  1. Final speed at the end of 5 seconds: [katex]\boxed{4.905, \text{m/s}}[/katex]
  2. Time to stop due to kinetic friction: [katex]\boxed{1.667, \text{seconds}}[/katex]
  3. Total time the block is in motion: [katex]\boxed{6.667, \text{seconds}}[/katex]

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Download Equation Sheet
KinematicsForces
\(\Delta x = v_i t + \frac{1}{2} at^2\)\(F = ma\)
\(v = v_i + at\)\(F_g = \frac{G m_1 m_2}{r^2}\)
\(v^2 = v_i^2 + 2a \Delta x\)\(f = \mu N\)
\(\Delta x = \frac{v_i + v}{2} t\)\(F_s =-kx\)
\(v^2 = v_f^2 \,-\, 2a \Delta x\) 
Circular MotionEnergy
\(F_c = \frac{mv^2}{r}\)\(KE = \frac{1}{2} mv^2\)
\(a_c = \frac{v^2}{r}\)\(PE = mgh\)
\(T = 2\pi \sqrt{\frac{r}{g}}\)\(KE_i + PE_i = KE_f + PE_f\)
 \(W = Fd \cos\theta\)
MomentumTorque and Rotations
\(p = mv\)\(\tau = r \cdot F \cdot \sin(\theta)\)
\(J = \Delta p\)\(I = \sum mr^2\)
\(p_i = p_f\)\(L = I \cdot \omega\)
Simple Harmonic MotionFluids
\(F = -kx\)\(P = \frac{F}{A}\)
\(T = 2\pi \sqrt{\frac{l}{g}}\)\(P_{\text{total}} = P_{\text{atm}} + \rho gh\)
\(T = 2\pi \sqrt{\frac{m}{k}}\)\(Q = Av\)
\(x(t) = A \cos(\omega t + \phi)\)\(F_b = \rho V g\)
\(a = -\omega^2 x\)\(A_1v_1 = A_2v_2\)
ConstantDescription
[katex]g[/katex]Acceleration due to gravity, typically [katex]9.8 , \text{m/s}^2[/katex] on Earth’s surface
[katex]G[/katex]Universal Gravitational Constant, [katex]6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2[/katex]
[katex]\mu_k[/katex] and [katex]\mu_s[/katex]Coefficients of kinetic ([katex]\mu_k[/katex]) and static ([katex]\mu_s[/katex]) friction, dimensionless. Static friction ([katex]\mu_s[/katex]) is usually greater than kinetic friction ([katex]\mu_k[/katex]) as it resists the start of motion.
[katex]k[/katex]Spring constant, in [katex]\text{N/m}[/katex]
[katex] M_E = 5.972 \times 10^{24} , \text{kg} [/katex]Mass of the Earth
[katex] M_M = 7.348 \times 10^{22} , \text{kg} [/katex]Mass of the Moon
[katex] M_M = 1.989 \times 10^{30} , \text{kg} [/katex]Mass of the Sun
VariableSI Unit
[katex]s[/katex] (Displacement)[katex]\text{meters (m)}[/katex]
[katex]v[/katex] (Velocity)[katex]\text{meters per second (m/s)}[/katex]
[katex]a[/katex] (Acceleration)[katex]\text{meters per second squared (m/s}^2\text{)}[/katex]
[katex]t[/katex] (Time)[katex]\text{seconds (s)}[/katex]
[katex]m[/katex] (Mass)[katex]\text{kilograms (kg)}[/katex]
VariableDerived SI Unit
[katex]F[/katex] (Force)[katex]\text{newtons (N)}[/katex]
[katex]E[/katex], [katex]PE[/katex], [katex]KE[/katex] (Energy, Potential Energy, Kinetic Energy)[katex]\text{joules (J)}[/katex]
[katex]P[/katex] (Power)[katex]\text{watts (W)}[/katex]
[katex]p[/katex] (Momentum)[katex]\text{kilogram meters per second (kgm/s)}[/katex]
[katex]\omega[/katex] (Angular Velocity)[katex]\text{radians per second (rad/s)}[/katex]
[katex]\tau[/katex] (Torque)[katex]\text{newton meters (Nm)}[/katex]
[katex]I[/katex] (Moment of Inertia)[katex]\text{kilogram meter squared (kgm}^2\text{)}[/katex]
[katex]f[/katex] (Frequency)[katex]\text{hertz (Hz)}[/katex]

General Metric Conversion Chart

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

  1. Start with the given measurement: [katex]\text{5 km}[/katex]

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

  3. Perform the multiplication: [katex]\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}[/katex]

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

Prefix

Symbol

Power of Ten

Equivalent

Pico-

p

[katex]10^{-12}[/katex]

Nano-

n

[katex]10^{-9}[/katex]

Micro-

µ

[katex]10^{-6}[/katex]

Milli-

m

[katex]10^{-3}[/katex]

Centi-

c

[katex]10^{-2}[/katex]

Deci-

d

[katex]10^{-1}[/katex]

(Base unit)

–

[katex]10^{0}[/katex]

Deca- or Deka-

da

[katex]10^{1}[/katex]

Hecto-

h

[katex]10^{2}[/katex]

Kilo-

k

[katex]10^{3}[/katex]

Mega-

M

[katex]10^{6}[/katex]

Giga-

G

[katex]10^{9}[/katex]

Tera-

T

[katex]10^{12}[/katex]

  1. 1. Some answers may vary by 1% due to rounding.
  2. Gravity values may differ: \(9.81 \, \text{m/s}^2\) or \(10 \, \text{m/s}^2\).
  3. Variables can be written differently. For example, initial velocity (\(v_i\)) may be \(u\), and displacement (\(\Delta x\)) may be \(s\).
  4. Bookmark questions you can’t solve to revisit them later
  5. 5. Seek help if you’re stuck. The sooner you understand, the better your chances on tests.

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