A constant force F, applied at an angle of 20° above the horizontal, pulls a 200 kg block, over a distance of 3 m, on a rough, horizontal floor as shown in the diagram below - NSC Physical Sciences - Question 5 - 2016 - Paper 1

The work-energy theorem states that the net work done on an object is equal to the change in kinetic energy of that object. In mathematical terms, it expresses that the work done is related to any change in kinetic energy.

In the free-body diagram, we identify the following forces acting on the block:

• The tension force ( F ) at an angle of 20° above the horizontal.
• The gravitational force ( W = mg ), acting downward.
• The normal force ( N ), acting upward from the surface.
• The frictional force ( F_f ), acting opposite to the direction of movement.

The work done by the kinetic frictional force can be expressed as: $W_k = f_{k} imes d = \\mu_k \times N \times d$

Where:

• ( \mu_k = 0.2 )
• The normal force ( N = mg - F \sin(20°) )

Thus, substituting values gives: ( W_k = 0.2 \times (mg - F \sin(20°)) \times 3 $$. Hence, rearranging leads to the expression provided.

To find the force ( F ) such that the net work done is zero, we consider: ( W_{net} = W_{input} - W_k = 0 ) Thus we set: $0 = W_{input} = [0] + (F \cos(20) \times 3) - W_k$ Substituting for ( W_k ) and solving provides: $F = 388.88 \text{ N}$