Question

A 2.12kg block initially at rest is pulled to the right along a horizontal surface by a constant horizontal force of 17.7 N. The coefficient of kinetic friction is 0.109. Find the speed of the block after it has moved 4.03 m.

Answer 1

Below

Step-by-step explanation:

Normal force = mg = 2.12 kg * 9.81 m/s^2 = 20.8 N

Ffriction = Fnormal * coeff of friction = 20.8 * .109 = 2.27 N

NET force moving block to right = 17.7 - 2.27 = 15.5 N

Remember F= m * a

15.5 = 2.12 * a shows a = 7.3 m/s^2

d = vo t + 1/2 a t^2 vo = 0 so this is just

d = 1/2 a t^2

4.03 = 1/2 ( 7.3 ) t^2 shows t = 1.05 s

then vf = at = 7.3 ( 1.05s) = 7.7 m/s