Question:
A body of mass 5kg has momentum of 10kgm/s.When a force of 0.2N is applied on it for 10s,find the change in its kinetic energy?
Answer:
Given m = 5 Kg, initial p = 10Kg m/s and F = 0.2 N for 10s
Let the initial momentum be p1 = mu and the corresponding Kinetic energy be K1
Let the final momentum be p2 = mv and the corresponding Kinetic energy be K2
p1 = mu = 10 = 5(u) Kg m/s ; Hence, u = 2 m/s
K1 = ½ mv2 = ½ (5) (2)2 = 10 J
P2 = mv
For finding v, we can use the second law of Newton – Force = Rate of change of momentum
F = dp/dt = m(v – u) / t
0.2 N = 5 (v -2) / 10 Hence, v = 12/5
P2 = mv = 5(12/5) = 12 Kg m/s
K2 = ½ (5) (12/5)2 = 14.4 J
Change in KE = K2 – K1 = 14.4 – 10 = 4.4 J
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