Question

A box of mass 6.0 kg is accelerated from rest

across a floor at a rate of 6.0 m/s² for 9.0 s.
Find the network done on the box

Answer 1

`\text{Given that },\\\text{Mass, m = 6 kg}\\\\\text{Acceleration, a = 6 } \text{m} \text{s}^(-2)\\\\\text{Time, t=9 sec}\\\\\text{Displacement, }s=ut + \frac 12 at^2 = \frac 12 at^2 = \frac 12 * 6 * 9^2 = 243 ~m}\\\\\ \text{Work,} ~W = Fs = mas = 6 * 6 * 243 = 8748 ~ J}`

Answer 2

Hi there!

We know that:

`W = \Delta KE = (1)/(2)m(\Delta v)^2`

We can begin by solving for the final velocity using the equation:

`v_f = v_i + at`

The box starts from rest, so:

`v_f = at\\\\v_f = 6(9) = 54 m/s`

Now, we can find the kinetic energy:

`KE = (1)/(2)mv^2\\\\KE = (1)/(2)(6)(54^2) = 8748 J`

Thus:

`W = \Delta KE\\\boxed{W = 8748 J}`