Problem 3- A block of mass m1 = 4 kg moves from rest under the influence of a varying force F(x) = 4x + 38, where F is in newton and x is in meters, on a rough horizontal surface having a coefficient of kinetic friction of 0.2. The block is connected by a light cord passing over a small frictionless pulley to a second hanging block of mass m2 = 3 kg as shown. If the system moves from x= 0 to x= 2m, calculate:

1) The work done by the force F(x) to move m1 from x= 0 to x= 2 m.
2) The change in the potential energy of each mass
3) The work done by friction.
4) The change in the kinetic energy of the system.

To solve this problem, we will need to consider the work done by the force F(x), the change in potential energy of each mass, the work done by friction, and the change in kinetic energy of the system. Let's break down each part:

1) The work done by the force F(x) to move m1 from x= 0 to x= 2m:
To calculate the work done by a force, we use the formula W = ∫F(x)dx, where F(x) is the force and dx is the displacement. In this case, the force F(x) = 4x + 38.
Integrating this with respect to x gives us:

W = ∫(4x + 38)dx
= 2x^2 + 38x | from x=0 to x=2

Substituting the limits, we get:

W = (2(2)^2 + 38(2)) - (2(0)^2 + 38(0))
= (8 + 76) - 0
= 84 J

Therefore, the work done by the force F(x) to move m1 from x= 0 to x= 2m is 84 Joules.

2) The change in the potential energy of each mass:
The change in potential energy is given by the formula ΔPE = mgh, where m is the mass, g is the acceleration due to gravity, and h is the change in height. In this case, m1 = 4 kg and m2 = 3 kg.

For m1:
ΔPE1 = m1 * g * h
= 4 kg * 9.8 m/s^2 * (2 m - 0 m)
= 78.4 J

For m2:
Since mass m2 is hanging, the change in height is given by the displacement x only.
ΔPE2 = m2 * g * h
= 3 kg * 9.8 m/s^2 * (0 m - 2 m)
= -58.8 J (negative because the change in height is downwards)

Therefore, the change in potential energy of m1 is 78.4 Joules and the change in potential energy of m2 is -58.8 Joules.

3) The work done by friction:
The work done by friction is given by the formula W_friction = μ * N * d, where μ is the coefficient of kinetic friction, N is the normal force, and d is the displacement.

The normal force N is equal to the weight of m1, which is N = m1 * g = 4 kg * 9.8 m/s^2 = 39.2 N.
The displacement d is 2 m.

W_friction = μ * N * d
= 0.2 * 39.2 N * 2 m
= 15.68 J

Therefore, the work done by friction is 15.68 Joules.

4) The change in kinetic energy of the system:
The change in kinetic energy is given by the formula ΔKE = W_net, where W_net is the net work done on the system.

In this case, W_net is equal to the work done by the force F(x) minus the work done by friction:
W_net = W_force - W_friction
= 84 J - 15.68 J
= 68.32 J

Therefore, the change in kinetic energy of the system is 68.32 Joules.