Newton's Second Law (Translational Dynamics) Practice

Sample questions

5 of 37 — sign in to practice the rest with adaptive difficulty and mastery tracking.

1. Sample 1difficulty 1/5

   A car cruises at constant $25~\text{m/s}$ on a level road. Which is true about the net force on the car?

   • A

     It is backward, equal to drag.

   • B

     It is downward, equal to gravity.

   • C

     It is forward, equal to the engine force.

   • D

     It is zero.

     check_circle

   Why

   Constant velocity → zero acceleration → zero <strong>net</strong> force. Engine force balances drag and friction.

   View full solution & statsarrow_forward

2. Sample 2difficulty 1/5

   A $3.0~\text{kg}$ object's velocity goes from $4~\text{m/s}$ east to $10~\text{m/s}$ east in $3.0~\text{s}$. The net force on it is

   • A

     $6~\text{N}$

     check_circle
   • B

     $2~\text{N}$

   • C

     $10~\text{N}$

   • D

     $14~\text{N}$

   Why

   $a = (10-4)/3 = 2~\text{m/s}^2$. $F = m a = 6~\text{N}$.

   View full solution & statsarrow_forward

3. Sample 3difficulty 1/5

   In an Atwood machine, the heavier side

   • A

     Accelerates downward.

     check_circle
   • B

     Depends on the rope length.

   • C

     Accelerates upward.

   • D

     Always stays still.

   Why

   Net torque/force from gravity tilts toward the heavier side, which descends.

   View full solution & statsarrow_forward

4. Sample 4difficulty 1/5

   A $2.0~\text{kg}$ object accelerates at $4.0~\text{m/s}^2$. What net force acts on it?

   • A

     $8.0~\text{N}$

     check_circle
   • B

     $0.5~\text{N}$

   • C

     $6.0~\text{N}$

   • D

     $2.0~\text{N}$

   Why

   $F_{\text{net}} = m a = (2.0)(4.0) = 8.0~\text{N}$.

   View full solution & statsarrow_forward

5. Sample 5difficulty 1/5

   A net horizontal force of $15~\text{N}$ accelerates a $5.0~\text{kg}$ crate on a frictionless surface. What is the crate's acceleration?

   • A

     $75~\text{m/s}^2$

   • B

     $3.0~\text{m/s}^2$

     check_circle
   • C

     $5.0~\text{m/s}^2$

   • D

     $0.33~\text{m/s}^2$

   Why

   $a = F_{\text{net}}/m = 15/5.0 = 3.0~\text{m/s}^2$.

   View full solution & statsarrow_forward