 # A 1.5 kg block is pushed to the right with just enough force to get it to move. The block is pushed for five seconds with this constant force, then the force is released and the block slides to a stop. If the coefficient of kinetic friction is 0.300 and the coefficient of static friction is 0.400. Calculate the amount of time that passes from when the force is applied to when the block stops.

6.67 seconds

## Written By Prof Phy AI

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

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

#### 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 F_{\text{net}} = F_{\text{push}} – F_{\text{kinetic}} Net force is the difference between pushing force and kinetic friction.
2 F_{\text{push}} = \mu_{s} F_{\text{normal}} Pushing force equal to maximum static friction.
3 F_{\text{kinetic}} = \mu_{k} F_{\text{normal}} Kinetic friction force.
4 F_{\text{net}} = \mu_{s} mg – \mu_{k} mg Substituting values for F_{\text{push}} and F_{\text{kinetic}}.
5 F_{\text{net}} = \mu_{s} mg – \mu_{k} mg = ma 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 v = u + \frac{F_{\text{net}}}{m}t Kinematic equation for velocity. Note F/m is the acceleration (1 m/s2 ) from above.
2 v = 0 + \frac{(\mu_{s} – \mu_{k}) mg}{m} \times 5, \text{s} Initial velocity u = 0; substituting F_{\text{net}} and t.
3 v = 5 m/s 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 0 = v – \mu_{k} gt_{\text{stop}} Final velocity 0 when stopped, v is final speed after push (5 m/s as found above).
2 t_{\text{stop}} = \frac{v}{\mu_{k} g} Solving for time to stop t_{\text{stop}}.

### Total Time

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

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

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

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6.67 seconds ## Suggest an Edit

##### Nerd-Notes.com
KinematicsForces
\Delta x = v_i \cdot t + \frac{1}{2} a \cdot t^2F = m \cdot a
v = v_i + a \cdot tF_g = \frac{G \cdot m_1 \cdot m_2}{r^2}
a = \frac{\Delta v}{\Delta t}f = \mu \cdot N
R = \frac{v_i^2 \cdot \sin(2\theta)}{g}
Circular MotionEnergy
F_c = \frac{m \cdot v^2}{r}KE = \frac{1}{2} m \cdot v^2
a_c = \frac{v^2}{r}PE = m \cdot g \cdot h
KE_i + PE_i = KE_f + PE_f
MomentumTorque and Rotations
p = m \cdot v\tau = r \cdot F \cdot \sin(\theta)
J = \Delta pI = \sum m \cdot r^2
p_i = p_fL = I \cdot \omega
Simple Harmonic Motion
F = -k \cdot x
T = 2\pi \sqrt{\frac{l}{g}}
T = 2\pi \sqrt{\frac{m}{k}}
ConstantDescription
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}
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 (kg·m/s)}
\tau (Torque)\text{newton meters (N·m)}
I (Moment of Inertia)\text{kilogram meter squared (kg·m}^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}}

Prefix

Symbol

Power of Ten

Equivalent

Pico-

p

10^{-12}

Nano-

n

10^{-9}

Micro-

µ

10^{-6}

Milli-

m

10^{-3}

Centi-

c

10^{-2}

Deci-

d

10^{-1}

(Base unit)

10^{0}

Deca- or Deka-

da

10^{1}

Hecto-

h

10^{2}

Kilo-

k

10^{3}

Mega-

M

10^{6}

Giga-

G

10^{9}

Tera-

T

10^{12}

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!