Question

A warehouse worker is pushing a 90.0 kg crate with a horizontal force of 298 N at a speed of v = 0.870 m/s across the warehouse floor. He encounters a rough horizontal section of the floor that is 0.75 m long and where the coefficient of kinetic friction between the crate and floor is 0.355.
Determine the magnitude and direction of the net force acting on the crate while it is pushed over the rough section of the floor.
Determine the net work done on the crate while it is pushed over the rough section of the floor.
Find the speed of the crate when it reaches the end of the rough surface.

Answer 1

Step 1 — friction force
N = mg = (90.0 kg)(9.80 m/s²) = 882 N
fk = μk N = 0.355(882 N) = 313.1 N (opposite the motion)

1) Net force
Fnet = Fpush − fk = 298 N − 313.11 N = −15.11 N
Magnitude = 15.1 N, direction: opposite to the push / opposite the crate’s motion.

2) Net work over 0.75 m
Wnet = Fnet · s = (−15.11 N)(0.75 m) = −11.33 J ≈ −11.3 J

3) Speed at end of rough patch (work–energy)
Wnet = ΔKE = 1/2 m(vf² − vi²)
vf² = vi² + 2Wnet/m = (0.870)² + 2(−11.33)/90.0
vf² = 0.7569 − 0.2518 = 0.5051
vf = √0.5051 = 0.711 m/s

Answers (3 s.f.):
- Net force: 15.1 N, opposite the motion.
- Net work: −11.3 J.
- Speed at end: 0.711 m/s.