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?

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,

Here,

= Kinetic coefficient

m = Mass

g = Gravitational acceleration

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

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

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

The work done by gravitational force is

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

Therefore the net work done by the system is 0J

connect the batteries in series and hope

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer