IIT-JEE 2010 · Paper 2 · Physics Q42

← IIT-JEE 2010 · Paper 2

PhysicsSingle correct+5 / -2Question 42 of 57

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

F(t) against t: a straight line falling from 4 N at t = 0, crossing zero at t = 3 s and continuing to −2 N at t = 4.5 s (marked by a dashed vertical line).

4.50 J

7.50 J

5.06 J

14.06 J

Answer and worked solution

Official answer: C

Impulse = area under the F-t graph. $F=4-\frac43t$ (zero at 3 s):
$\int_0^{4.5}F\,dt=\frac12(4)(3)-\frac12(2)(1.5)=6-1.5=4.5\ \mathrm{N\,s}.$

Kinetic energy. Starting from rest, $p=4.5$ kg m s⁻¹:
$K=\frac{p^2}{2m}=\frac{20.25}4=5.06\text{ J}.$

(C)

Topics:Impulse and momentum Work-energy theorem

← Previous question Next question →

IIT-JEE 2010 · Paper 2, Physics question 42