A 200 kg piano is moved 6.8 m across a smooth floor by a force of 220 N directed downward at 25 degrees.

Question

A 200 kg piano is moved 6.8 m across a smooth floor by a force of 220 N directed downward at 25^o

W = F·cos(𝜃)·d

W_net = ΔK = K_Final − K_Initial

_{K=1/2 mv^2}

A. How much work is done by the force on the piano?

W =  J

B. What will the piano's kinetic energy be upon being moved?

K =  J

C. What will the piano's final speed be?

v =  m/s

I was able to solve part A by using the equation W = F·cos(𝜃)·d and plugged in the numbers 220N, cos (25), 6.8 to get 1.355 e3 J. For the next two parts I am not sure how to start. :(

Answer 1

Solution:

A) The first section is correct as you solved it

B) The floor is smooth meaning there are no friction. Applying newton 2nd law in the Horizontal direction

∑F = ma , the only forces are the traction force F = 220 N, a = F/m = 220N / 200kg = 220 kg.m/s² / 200kg

a = 1.1 m/s² . The velocity can be obtained from the equations of motion. knowing that the piano starts from rest then x_o = v_o = 0

using the equation x = 1/2at²+v_o t + x_o , x_o = v_o = 0 , 6.8m = 1/2 at², at² = 6.8m(2) = 13.6 m, and t² = 13.6m/a = 13.6m/1.12m/s² = 12.36 s² , and t = √12.36 s² = 3.51 s

from the equation v_f = at + v_o , v_f = 1.1 m/s²( 3.51s) = 3.86 m/s

The kinetic Energy is KE = 1/2 m v² = 1/2 ( 200kg)(3.86m/s)² = 1490 kg.m²/s² = 1490 N.m = 1490 J

C) The final velocity at the end of the 6.8 m is v_f = 3.86 m/s