Question

1a) A rope is used to pull a box across a friction-less surface. The rope is held at an angle of 40 degrees above the horizontal, and a force of 650 N is applied, resulting in a horizontal displacement of 15 m. How much work does the force on the rope do?

1b) Assume the same set of conditions as the above problem, but this time the box is pulled across a surface where a 80 N force of friction acts against the box. How much work is done in this case?

Please provide a complete explanation and answer

Answer 1

(a)work is done in the horizontal direction as the displacement is in horizontal directon.
work=component of force in the direction of displacement x displacement
work=650 cos40x15

Answer 2

(b)the external force is opposed by frictional force.so the net force acting on the body in the horizontal direction is reduced(frictional force is acting parallel to the plane).
work=net force applied to move the object x displacement
work=(650 cos40-80)x15

Answer 3

If the applied force and its angle and the distance are the same, then the work done is the same, just that some of it has gone into overcoming friction. The first case was rather artificial, as continuing the same force on a frictionless surface would produce continuous acceleration of the object.

Answer 4

I'm with Douglas on this - the question asks about the work done by the Force in the rope.
In the first case, with no frictional losses, ALL the work would be converted to Kinetic Energy of the box.
In the second case - the frictional force does work in the opposite direction (-15m x80N) and this amount of energy is therefore 'lost'. The box therefore accelerates across the surface more slowly, and hence has less KE at the end of the 15m travel.