g A dockworker loading crates on a ship finds that a 22-kg crate, initially at rest on a horizontal surface, requires a 72-N horizontal force to set it in motion. However, after the crate is in motion, a horizontal force of 52 N is required to keep it moving with a constant speed. Find the coefficients of static and kinetic friction between crate and floor.

μs = 0.333 and μk = 0.24

Explanation:

Mass of crates = 22kg

F = 72newton

F =μR

Also given that g = 9.8m/s^2 which is that acceleration due to gravity

Hence R = N = mg

= 9.8 × 22

= 215.6 newton

F is frictional force

μ is coefficient of friction

N or R is normal reaction

Hence, the static friction

F =μsR

72 = μs × 215.6

μs = 72/215.6

μs = 0.333

The kinetic friction coefficient

F =μkR

When F = 52N and R = N = 215.6N

52 = μk × 215.6

μk = 52/215.6

μk = 0.24

1.195000

explanation:

total mass=0.0117+0.0137+0.0881+0.0111

total mass= 0.01172kg

total velocity = 0.283-0.743+0.871-0.799+0.155+0.637+0.301=0.102

momentum=mv=0.1172*0.102=1.195000//