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

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

Answer 2

Hi there!

We know that:

$W = \Delta KE = \frac{1}{2}m(\Delta v)^2$

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:

$KE = \frac{1}{2}mv^2\\\\KE = \frac{1}{2}(6)(54^2) = 8748 J$

Thus:

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