Question

If the coefficient of kinetic friction between the 150lb crate and the ground is us=0.21. Determine the speed of the crate when t=4s. The crate starts from rest and is towed by the 100lb force.

Answer 1

Solution:

Note:

1 lb = 0.45 kg

m = 150 lb = 68 kg

F = 100 lb = 45 kg = 45 x 9.8 N = 441 N

Thus by: F(net) = F(applied) - F(friction)

ma = F - μk * N

68 * a = 441 - 0.21 * (mg)

68 * a = 441 - 0.21 * 68 * 9.8

a = 4.43 $( \tfrac{m}{s^{2}})$

Thus by: $v=v_{o}+at$

$v$ = 0 + 4.43 * 4

$v$ = 17.71 $( \tfrac{m}{s})$