Answer:
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
Answer:
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.
