If you push a crate horizontally with 100 N across a 10-m factory floor and the friction between the crate and the floor is a steady 70 N, how much kinetic energy does the crate gain?

300 J.

Explanation:

kinetic Energy: This can be defined as the energy of a body by virtue of it's motion. The S. I unit of kinetic energy is Joules (J).

From the question, Applying the law of conservation of energy,

Ek = F'*d ... Equation 1

Ek = kinetic energy of the crate, F' = resultant force on the crate, d = distance moved by the crate.

But,

F' = F-Fr ... Equation 2

Where F = Horizontal force on applied to the crate, Fr = Friction force

Substitute equation 2 into equation 1

Ek = (F-Fr) d ... Equation 3

Given: F = 100 N, Fr = 70 N, d = 10 m.

Substitute into equation 3

Ek = (100-70) 10

Ek = 30 (10)

Ek = 300 J.

300J

Explanation:

Parameters given:

Force applied on crate, F = 100N

Distance moved by crate, D = 10m

Frictional force, Fr = 70N

The work done by pushing the crate is given by:

W = F * D

= 100 * 10 = 1000J

The work done by friction is given as:

W (Fr) = - Fr * D

= - 70 * 10 = - 700J

Work done by friction is negative because it works against the motion of the crate.

Net work done = 1000 + (-700)

WD = 1000 - 700 = 300J

Work done is equivalent to the change in kinetic energy.

WD = ∆KE

∆KE = KE (final) - KE (initial)

KE (initial) = 0

Therefore,

WD = KE (final)

=> KE (final) = 300J