Question

A factory worker moves a 30.0 kg crate a distance of 4.5 m along a level floor at constant velocity by pushing horizontally on it. The coefficient of kinetic friction between the crate and the floor is 0.25.

1. What magnitude of force must the worker apply?
2. How much work is done on the crate by the worker's push?
3. How much work is done on the crate by friction?
4. How much work is done by normal force? By gravity?
5. What is the net work done on the crate?

Answer 1

To solve this problem we will apply the concepts related to the Friction force and work. The friction force can be defined as the product between the Normal Force (Mass by gravity) and the dynamic friction constant. In the case of Work this is defined as the product of the distance traveled by the applied force. Then we will solve the points sequentially to find the answer to each point,

PART A) The friction force with the given data is,

`F_f = \mu_k mg`

Here,

`\mu_k` = Kinetic coefficient

m = Mass

g = Gravitational acceleration

`F_f = (0.25)(30kg)(9.8m/s^2)`

`F_f = 73.5N`

PART B) The work done by the worker is the distance traveled for the previously force found, then

`W_w = rF_f`

`W_w = (4.5m)(73.5N)`

`W_w = 330.75J`

PART C) The work done by the friction force would be the distance traveled with the previously calculated force, therefore

`W_f = -rF_f`

`W_f = -(4.5m)(73.5N)`

`W_f = -330.75J`

`W_f = -331J`

PART D) The work done by the normal force is,

`W_N = N (r) Cos(90)`

`W_N = 0J`

The work done by gravitational force is

`W_g = rF_gcos(90)`

`W_g = 0J`

PART E) The expression for the total work done is,

`W_(net) = W_f +W_w +W_N+W_g`

`W_(net) = 330.75-330.75+0+0`

`W_(net) = 0J`

Therefore the net work done by the system is 0J