A rope is used to pull a 3.57 kg block at constant speed 4.06 m along a horizontal floor. The force on the block from the rope is 7.68 N and directed 15.0° above the horizontal. What are (a) the work done by the rope's force, (b) the increase in thermal energy of the block-floor system, and (c) the coefficient of kinetic friction between the block and floor?

A) W = 30.12J

B) Δthermal = 30.12 J

Explanation:

We are given that;

Mass; m = 3.57 kg

Force on block; F = 7.68N

θ = 15°

d = 4.06m

The work done by the rope forces will be;

W = Fd cosθ = 7.68 x 4.06 x cos 15 = 30.12J

B) The block moves with constant speed and so the pulling force will be equal to the friction force.

Thus;

the increase in thermal energy of the block-floor system will be equal to work done; thus;

Δthermal = 30.12 J

C) To calculate this, resolving forces, we obtain;

Fcos θ = μ (mg - F sin θ)

Where μ is coefficient of kinetic friction.

Thus, making μ the subject, we have;

Fcosθ / (mg - F sin θ) = μ

Plugging in relevant values to obtain;

μ = 7.68cos15 / (3.57x9.8 - 7.68 sin 15) = 7.4183 / (34.986 - 1.9877) = 0.225