Question

•A winch is used to slide a 350-kg load at constant speed 3.50 m down a 27.0° incline. The rope between the winch and the load is parallel to the incline. The coefficient of kinetic friction between the ramp and the load is 0.40. Calculate the force exerted by the winch, the work done by the winch on the load, the work done by the friction force, the work done by the force of gravity, and the net work done on the load.

Answer 1

Since the net force on the load is zero (As, acceleration is zero)

Fx - Ff + Fa = 0

The force applied by the winch will be:

Fa = -Fx + Ff

here, Fx = friction force by gravity in inclined direction = m*g*sin = 350*9.81*sin(27.0 deg)

Ff = friction force acting on the load = *(Normal force(N)) =

Ff = *Fy = *m*g*cos = 0.40*350*9.81*cos(27.0 deg)

So, Fa = force applied by winch = -350*9.81*sin(27.0 deg) + 0.40*350*9.81*cos(27.0 deg)

Fa = -335 N

now,

The work done by the winch on load will be:

Wa = Fa*d

here, d = displacement = 3.5 m

then, Wa = -335*3.5 = -1172.5

Wa = -1173 J

the work done by friction will be:

Wf = -Ff*d = -0.40*350*9.81*cos(27.0 deg)*3.5

Wf = -4282.93 J

Wf = -4283 J

the work done by gravity is

Wg = F*d

here, F = gravity force = m*g = 350*9.81

d = vertical displacement = 3.50*sin(27.0 deg)

then, Wg = 350*9.81*sin(27.0 deg)*3.50 = 5455.72

Wg = 5456 N

finally, the net work done on the load will be:

Wnet = Wa + Wf + Wg = Fd

Wnet = -1173 - 4283 + 5456

Wnet = 0 J