A constant force of 5 N acts along the direction of motion on a 2 kg block for 4 m on a frictionless surface. How much work is done?

A constant force of 5 N acts along the direction of motion on a 2 kg block for 4 m on a frictionless surface. How much work is done?

• A. 10 J
• B. 4 J
• C. 20 J
• D. 40 J

Answer: (C)

Work is the energy transferred by a force as it acts through a displacement. For a constant force, the work is W = F d cos(theta). The force here points in the same direction as the motion, so theta = 0 and cos(0) = 1. Thus the work done is W = 5 N × 4 m = 20 J. On a frictionless surface, this work goes entirely into increasing the block’s kinetic energy, with none dissipated as heat. The mass isn’t needed to compute the work itself; it would affect acceleration via F = ma, but not the amount of work for this displacement.

Work is the energy transferred by a force as it acts through a displacement. For a constant force, the work is W = F d cos(theta). The force here points in the same direction as the motion, so theta = 0 and cos(0) = 1. Thus the work done is W = 5 N × 4 m = 20 J. On a frictionless surface, this work goes entirely into increasing the block’s kinetic energy, with none dissipated as heat. The mass isn’t needed to compute the work itself; it would affect acceleration via F = ma, but not the amount of work for this displacement.