Question

The relation between the displacement x and the
time t for a body of mass 2 kg moving under the
action of a force is given by x = ť°/3, where x is in
metre and t in second, calculate the work done by
the body in first 2 seconds.
(Ans. 16 )​

Answer 1

16 J is the required answer.

GIVEN:

• Displacement, x = t³/3
• Mass of the body, m = 2 kg
• Time, t = 2 seconds.

TO FIND:

Work done by the body, w

FORMULAE:

• Differentiation of displacement gives instantaneous velocity. v = dx/dt

• Differentiation of velocity gives instantaneous acceleration. a = dv/dt

• According to Newton's second law of motion, f = ma

• Work done, dw = f • dx

SOLUTION:

STEP 1: TO FIND VELOCITY

$v = \dfrac{dx}{dt} \\ \\ v = \dfrac{d}{dt} ( \dfrac{ {t}^{3} }{3} ) \\ \\ v = \dfrac{1}{3} \dfrac{d}{dt} ( {t}^{3} ) \\ \\ v = \dfrac{1}{3} \times 3 \times {t}^{2} \\ \\ v = {t}^{2}$

STEP 2: TO FIND ACCELERATION

$a = \frac{dv}{dt} \\ \\ a = \frac{d}{dt} ( {t}^{2} ) \\ \\ a = 2t$

STEP 3: TO FIND FORCE

$f = ma \\ \\ f = 2 \times 2t \\ \\ f = 4t$

STEP 4 : TO FIND dx

$x = \dfrac{ {t}^{3} }{3} \\ \\ \dfrac{dx}{dt} = \dfrac{3 \times {t}^{2} }{3} \\ \\ \frac{dx}{dt} = {t}^{2} \\ \\ dx = {t}^{2} dt$

STEP 5: TO FIND WORK

$dw = f.dx \\ \\ dw = 4t \times {t}^{2} dt \\ \\ dw = 4 {t}^{3}dt \\ \\ w = \int\limits_{0}^{2}4 {t}^{3} dt \\ \\ w = 4\int\limits_{0}^{2} {t}^{3} dt \\ \\ w = 4 \times \sf{\bigg[{\dfrac{t^4}{4}}\bigg]^{2}_o} \\ \\ w = \frac{4}{4} \times ( {2}^{4} - {0}^{4} ) \\ \\ w = 1 \times {2}^{4} \\ \\ w = 16 \: joule$

ANSWER:

Work done by the body during 2 seconds is 16 J.