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

34304 Joule

Step-by-step 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* \int _(0)^(8) \left ( -24t+82 t - 72t^(2)+18t^(3) \right )dx`

`W=4* \left ( -24t+41t^(2) - 24t^(3)+4.5t^(4)\right )_(0)^(8)`

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

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

W = 34304 Joule

Answer 2

31.232 kJ

Step-by-step 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 = (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 = (1)/(2)mv^2 - (1)/(2)mu^2`

`W = (1)/(2)m(v^2 -u^2)`

`W = (1)/(2)* 4* (125^2 -3^2)`

W = 31.232 k J

Hence, work done is equal to 31.232 kJ