A 9 kg cart is pulled by a force of 16 newtons for a distance of 6 meters, then by a force of 7 Newtons for 2 meters, then it is pushed (slowed down) by a force of 5 Newton for 3 meters. All this occurs on a horizontal surface.

a) Initially assuming the surface to be frictionless, what net work is done on the cart as it moved the 11 meters?

b) If the cart starts from rest, what is its speed after it has moved the 11 m?

Now assume a coefficient of friction µ between cart and surface during the entire distance.
c) For what value of μ does the cart just come to a halt after 11 m?

d) For the value of µ computed in part (c), what is the speed of the cart after it has gone 4 m?

My Answer™ :~

Applying work - energy theorem, we get :-

w_f1 + w_f2 + w_f3 = 16*6 + 7*2 + 5*3 = 125 J

 Ans (A) = 125 J

W = change in kinetic energy

hence,

125 = .5 * 9 * v²

v = 5.27 m/s

Ans (B) = 5.27 m/s

for the car to come to a holt , final kinetic energy = 0

125 - Nd  = 0 - 0

125 = 90 * 11

= .12

Ans (C) :~   = .12

W_f1 - W_friction = Final Kinetic Energy

96 - .12 * 90 * 4 = .5 * 9 * v²

96 - 43.2 = 4.5v²

v = 3.42  m/s

Ans (D) :~  v = 3.42 m/s