A 4000 kg tractor is pulling a 1500 kg cart The tractor push

A 4000 kg tractor is pulling a 1500 kg cart. The tractor pushes backwards on the ground with a force of 3600N. Therefore the ground pushes back equally with a force of 3600 N. The tractor and the cart accelerate forward at a rate of 0.4 m/sec^2. What is the tension in the rope? What is the force of friction on the cart? What is the coefficient of friction for the cart?

Solution

a) Tension in the rope, T = 3600 N

b) net force acting on cart, Fnet = T - Friction

m_cart*a = T - Friction

Friction = T - m_cart*a

= 3600 - 1500*0.4

= 3000 N

c)let mue_k is the coefficient of friction.

we know, kinetic friction = mue_k*N

3000 = mue_k*m*g

mue_k = 3000/(m*g)

= 3000/(1500*9.8)

= 0.204