Parker (73.2 kg) is being dragged down the hall with an applied force of 123 N. If the frictional force is 27.4 N, what is the coefficient of friction in the hall?

The coefficient of friction in the hall is 0.038

Explanation:

Given;

mass of the Parker, m = 73.2 kg

applied force on the parker, F = 123 N

frictional force, Fs = 27.4 N

the coefficient of friction in the hall = ?

frictional force is given by;

Fs = μN

Where;

μ is the coefficient of friction

N is normal reaction = mg

Fs = μmg

μ = Fs / mg

μ = (27.4) / (73.2 x 9.8)

μ = 0.038

Therefore, the coefficient of friction in the hall is 0.038

scientific laws are the result of testing theories and proving them to be facts.

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//