A worker pushes 25Kg crate a distance of 5m along a level floor at constant velocity by pushing horizontally on it. The coefficient of kinetic frication between the crate and the floor is 0.23.

(a) What magnitude of force must the worker apply?
(b) How much work is done on the crate by this force?
(c) How much work is done on the crate by frication?
(d) How much work is done on the crate by normal force?
(e) How much work is done on the crate by the gravity?
(f) What is the total work done on the crate?

(a) The applied force must equal the friction force. You should know how to calculate that, using the friction coefficient.

(b) Force x (5 m)
(c) minus (b) answer
(d) zero
(e) zero
(f) zero

The answer above is very stupid

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way to go man, because if I didn't know the equations, your snarkiness will show me the light.

(a) To find the magnitude of force the worker must apply, we can use Newton's second law of motion which states that the net force on an object is equal to the mass of the object multiplied by its acceleration. In this case, since the crate is moving at a constant velocity, its acceleration is zero.

The force of friction can be calculated using the formula: frictional force = coefficient of friction * normal force.

Since the crate is on a level floor, the normal force is equal to the weight of the crate, which can be calculated using the formula: weight = mass * gravity.

Let's calculate the force:

Mass of the crate = 25 kg
Gravity = 9.8 m/s^2 (acceleration due to gravity on Earth)
Coefficient of kinetic friction = 0.23

Weight of the crate = mass * gravity = 25 kg * 9.8 m/s^2 = 245 N
Force of friction = 0.23 * 245 N = 56.35 N

Since the crate is moving at constant velocity, the worker must apply a force equal in magnitude but opposite in direction to the force of friction. Therefore, the magnitude of the force the worker must apply is 56.35 N.

(b) The work done on an object is equal to the force applied on the object multiplied by the displacement of the object in the direction of the force. Since the crate is moved over a distance of 5m in the horizontal direction, the worker applies a force and does work on the crate.

Work done = force applied * displacement

Work done = 56.35 N * 5m = 281.75 N*m or Joules

Therefore, the work done on the crate by the worker is 281.75 Joules.

(c) The work done on the crate by friction can be calculated using the formula: work done = force of friction * displacement.

Since the crate is being pushed horizontally along the floor, the displacement is 5m.

Work done by friction = force of friction * displacement

Work done by friction = 56.35 N * 5m = 281.75 N*m or Joules

Therefore, the work done on the crate by friction is 281.75 Joules.

(d) The work done on the crate by the normal force is zero. This is because the normal force is perpendicular to the displacement of the crate, and work is only done when a force acts in the same direction as the displacement.

Therefore, the work done on the crate by the normal force is zero.

(e) The work done on the crate by gravity can be calculated using the formula: work done = weight * displacement.

Since the crate is moved horizontally along the floor, there is no vertical displacement, and therefore, no work is done on the crate by gravity.

Therefore, the work done on the crate by gravity is zero.

(f) The total work done on the crate is the sum of the work done by the worker (281.75 Joules), the work done by friction (281.75 Joules), and the work done by gravity (zero Joules).

Total work done on the crate = Work done by the worker + Work done by friction + Work done by gravity
Total work done on the crate = 281.75 Joules + 281.75 Joules + 0 Joules = 563.5 Joules

Therefore, the total work done on the crate is 563.5 Joules.