Certain force acting on a 20 kg mass changes its velocity from 5 ms-1 to 2 ms-1.Calculate the work done by the force.
Answers
Given : Initial velocity u=5 m/s
Final velocity v=2 m/s
The work done is equal to the change in the kinetic energy of an object.
W=
2
1
m(v
2
−u
2
)
W=
2
1
×20×(2
2
−5
2
)
⟹W=−210 J
Answer:
The work done by the force is -210 J.
Explanation:
A force acts on the body and changes its velocity. By the work-energy theorem, the work done will be equal to the change in kinetic energy of the system.
The mass of the body, m = 20 Kg
The initial velocity of the body, u = 5 m/s
Final velocity of the body, v = 2 m/s
The equation to find the kinetic energy is,
K.E = (mv²) / 2
The initial kinetic energy of the body, K.E₁ = (mu²) / 2
= (20 × 5²) / 2
= 500 / 2
= 250 J
The final kinetic energy of the body, K.E₂ = (mv²) / 2
= (20 × 2²) / 2
= 80 / 2
= 40 J
Work done by the force = Change in kinetic energy
= K.E₂ - K.E₁
= 40 - 250
= -210 J
The negative sign indicates that the work is done by force against the direction of motion of the body.