Question

A single force acts on a 4.0 kg particle-like object in such a way that the position of the object as a function of time is given by x = 3.0t − 4.0t 2 + 1.0t 3, with x in meters and t in seconds. Find the work done on the object by the force from t = 0 to t = 8.0 s.

Answer 1

Answer:

31.232 kJ

Explanation:

Given,

mass of the particle, m = 4 Kg

Position of object as the function of time

$x = 3.0 t - 4.0 t^2 + 1.0 t^3$

Work done by the object by the force from t = 0 to t = 8.0 s =?

we know,

$v = \dfrac{dx}{dt}$

$v =3-8t +3 t^2$

velocity at t = 0 s

u = 3 - 8 x 0 + 3 x 0 = 3 m/s

velocity at t= 8 s

v = 3 - 8 x 8 + 3 x 8 x 8

v = 125 m/s

Work done is equal to change in KE

$W = \dfrac{1}{2}mv^2 - \dfrac{1}{2}mu^2$  

$W = \dfrac{1}{2}m(v^2 -u^2)$

$W = \dfrac{1}{2}\times 4\times (125^2 -3^2)$

W = 31.232 k J

Hence, work done is equal to 31.232 kJ

Answer 2

Answer:

34304 Joule

Explanation:

mass of particle, m = 4 kg

x = 3t - 4t² + t³

Let v is the velocity

v = dx/dt = 3 - 8t + 3t²

Let a is the acceleration

a = dv/dt = - 8 + 6t

Work is defined as the product of force.

$\int dW=\int madx$

$W=4\times \int _{0}^{8} \left ( -24t+82 t - 72t^{2}+18t^{3} \right )dx$

$W=4\times \left ( -24t+41t^{2} - 24t^{3}+4.5t^{4}\right )_{0}^{8}$

$W=4\times \left ( -24\times 8+41\times 64 - 24\times 512+4.5\times 4096\right )$

W = 4 x (- 192 + 2624 - 12288 + 18432)

W = 34304 Joule