A horizontal force of 12 newtons is applied to a 4.0 kg box that slides on a horizontal surface. The box starts from rest, moves a horizontal distance of 10 meters, and obtains a velocity of 5.0 m/s. The surface has friction. The friction force, assumed constant, is
A) 7.0 N.
B) 6.5 N.
C) 6.0 N.
D) 5.7 N.
E) 4.9 N.

Question

Quizplus · Accepted Answer

The work done by the applied force is equal to the change in kinetic energy plus the work done against friction. The change in kinetic energy (ΔKE) can be calculated using the formula ΔKE = 0.5 * m * v^2, where m is the mass and v is the velocity. For a 4.0 kg box moving at 5.0 m/s, ΔKE = 0.5 * 4.0 * (5.0)^2 = 50 J. The work done by the applied force (W) is force * distance = 12 N * 10 m = 120 J. The work done against friction (W_f) is the difference between these two, W - ΔKE = 120 J - 50 J = 70 J. Since this work is done over a distance of 10 meters, the friction force (F_f) can be found by rearranging the work formula W_f = F_f * distance, giving F_f = W_f / distance = 70 J / 10 m = 7.0 N.

A Horizontal Force of 12 Newtons Is Applied to a 4.0