Friction Physics Example 2

Follow the full solution, then compare it with the other examples linked below.

Example 2

hard
A 15 kg15 \text{ kg} block on a surface (μk=0.25\mu_k = 0.25) is pushed with 80 N80 \text{ N}. What is the acceleration?

Solution

  1. 1
    Normal force: N=mg=15×9.8=147 NN = mg = 15 \times 9.8 = 147 \text{ N}.
  2. 2
    Friction force: fk=μkN=0.25×147=36.75 Nf_k = \mu_k N = 0.25 \times 147 = 36.75 \text{ N}.
  3. 3
    Net force: Fnet=80−36.75=43.25 NF_{\text{net}} = 80 - 36.75 = 43.25 \text{ N}.
  4. 4
    Acceleration: a=Fnetm=43.2515≈2.88 m/s2a = \frac{F_{\text{net}}}{m} = \frac{43.25}{15} \approx 2.88 \text{ m/s}^2

Answer

a≈2.88 m/s2a \approx 2.88 \text{ m/s}^2
When a force exceeds friction, the object accelerates. The net force is the applied force minus friction, and the acceleration follows from Newton's second law.

About Friction

A contact force that opposes the relative motion or tendency of motion between two surfaces in contact.

Learn more about Friction →

More Friction Examples

Example 1 medium

A [formula] box sits on a surface with a coefficient of kinetic friction [formula]. What force is ne

Example 3 medium

The coefficient of static friction between a box and the floor is [formula]. If the box weighs [form

Example 4 medium

A [formula] crate is pulled across a level floor at constant speed by a horizontal [formula] force.

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