A block of mass 2 kg is free to move along the x-axis. It is at rest and from t=0 onwards it is subjected to a time-dependent force F(t) in the x direction. The force F(t) varies with t as shown in the figure. The kinetic energy of the block after 4.5 seconds is
(A)4.50 J
(B)7.50 J
(C)5.06 J✓ Correct
(D)14.06 J
Answer: (C)
Solution
The impulse equals the change in momentum:
∫Fdt=Δp=mv
The area under the F-t curve from t = 0 to t = 4.5 s gives the impulse. From the graph, this area is 4.5 kg·m/s.
Therefore, the final velocity is:
v=24.5=2.25 m/s
The kinetic energy is:
KE=21mv2=21(2)(2.25)2=5.06 J
The answer is (C) 5.06 J.
JEE Advanced · 2010 · 11 Apr 2010 · Paper 2 · Single correct option