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. (3 marks)

b) The net force acting on the piano. (3 marks)

c) The normal force acting on the piano. (3 marks)

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

Answer 1

a) We can find the acceleration of the piano from the formula:

$$a=\dfrac{v-v_o}{t}=\dfrac{0.92\ \dfrac{m}{s}-0}{5.0\ s}=0.184\ \dfrac{m}{s^2}.$$

b) We can find the net force acting on the piano from the Newton's Second Law of Motion:

$$F_{net}=ma=275\ kg\cdot0.184\ \dfrac{m}{s^2}=50.6\ N.$$

b) We can find the normal force acting on the piano from the formula:

$$F_N=mg=275\ kg\cdot9.8\ \dfrac{m}{s^2}=2695\ N.$$

d) Let's apply the Newton's Second Law of Motion:

$$F_{push}-F_{fr}=ma,$$ $$F_{push}-\mu_k mg=ma,$$ $$\mu_k=\dfrac{F_{push}-ma}{mg},$$ $$\mu_k=\dfrac{1075\ N-275\ kg\cdot 0.184\ \dfrac{m}{s^2}}{275\ kg\cdot 9.8\ \dfrac{m}{s^2}}=0.38$$