Question

A worker pushed a 21 kg block 8.0 m along a level floor at constant speed with a force directed 30° below the horizontal. If the coefficient of kinetic friction between block and floor was 0.30, what were (a) the work done by the worker's force and (b) the increase in thermal energy of the block-floor system?

Answer 1

This is a vector problem. We start with a force vector, magnitude = F, direction 30 degrees down from the horizontal. Since the force is not aligned with the direction of motion, the first thing you want to do is break it down into components along the direction of motion (horizontal) and perpendicular to it (vertical).

The horizontal component is given by the worker's force times the cosine of 30°, or 0.8660F.
The vertical component is given by the worker's force times the sine of 30°, or 0.500F.

The block is moving at a constant speed, so the acceleration is zero. This means the *net* force is precisely zero. The horizontal force applied by the worker is exactly matched by the force of friction.

The resistive force due to friction is equal to the coefficient of kinetic friction times the weight of the block plus the vertical component of the force.
Weight is given as m*g = 21 * 9.81 = 206.01N
R = µ(206.01kg + 0.500F)

Because the net force is zero, the resistive force is equal to the horizontal component of the force exerted by the worker:
R = 0.30 * (201.06kg + 0.5F) = 0.8660F
= 60.31N + 0.15F = 0.866F
A bit more algebra gives us
60.31N = 0.866F - 0.15F = 0.716F
F = 60.31N / 0.716= 84.23N

The worker is pushing with a force of 84.23 Newtons.
(Check: 84.23 *0.866 = 72.94; 84.23 * 0.5 = 42.115; 0.30 * (201.06+42.115) = 72.95--- well within rounding error.)

The amount of work done is the horizontal component of the force multiplied by the distance along which it is exerted: 72.95 N * 8M = 583.62.J
Since the block is not accelerating, all the work done is being used up overcoming friction, and thus all of it goes into thermal energy. The increase in thermal energy is thus 583.62J.

Since one calorie is 4.18J, this means 139.62 calories have been imparted to the block-floor system