A body of mass 5 kg is moving with a momentum of 10 kg metre per second a force of 0.2 in the direction of motion of body for 10 second the increase in kinetic energy is what?
Answers
Solution :
Given,
Mass, m = 5kg
Momentum, p = 10 kg m/s
Force, F = 0.2N
Time, t = 10s
Suppose, v1 be initial velocity and v2 be final velocity.
For finding the initial velocity
p = mv
=> p = mv1
=> p = 5 × v1
=> 10 = 5 × v1
=> 10/5 = v1
=> v1 = 10/5 =2
=> v1 = 2m/s
.°. v1 = 2m/s
For finding the initial velocity
F = m × a
=> F = m × a
=> 0.2 = 5 × a
=> 0.2 = 5a
=> 5a = 0.2
=> a = 0.2/5
=> a = 0.04m/s^2
.°. a = 0.04m/s^2
For finding the final velocity
v = u + at
=> v = u + at
=> v = 2 + 0.04 × 10
=> v = 2.4m/s
.°. v2 = 2.4m/s
The formula of kinetic energy :
K.E. = 1/2 mv^2
By using the formula of change in the kinetic energy.
Change in Kinetic Energy = 1/2mv2^2 - 1/2mv1^2
=> Change in Kinetic Energy = 1/2mv2^2 - 1/2mv1^2
=> Change in Kinetic Energy = 1/2 × 5 × (2.4)^2 - 1/2 × 5 × (2)^2
=> Change in Kinetic Energy = 14.4J - 10J = 4.4J
=> Change in Kinetic Energy = 4.4J
We know that,
Change in kinetic energy = Increase in kinetic energy.
.°. Change in Kinetic Energy = 4.4J
.°. Increase in Kinetic Energy = 4.4J