Question

A piano mover pushes a piano (m = 275 kg) with an applied horizontal of 1075 N. This causes the piano to accelerate, from rest, to a final velocity of 0.92 m/s over a time of 5.0 s. Calculate:

a) The acceleration of the piano.

b) The net force acting on the piano.

c) The normal force acting on the piano.

d) The coefficient of kinetic friction between the piano and the floor.

Answer 1

Solution:

acceleration: $a=\dfrac{ΔV}{Δt}=\dfrac{0.92-0}{5-0}=0.184\;(\tfrac{m}{sec^2})$

motion formula :

m*(g*μ k+a) = F

275*(9.806*μ k+0.184) = 1075 

2700*μ k = 1075-2700*0.184 = 580

friction coefficient μ_k = $\dfrac{29}{135}$ = 0.215

net force NF = m*a = 275*0.184 = 50.6 N

normal force N = m*g = 275*9,806 = 2700 N