Question

help on #5 pls

the answer is 560 J, just not sure how to get it

5. A 5.4 kg box is initially at rest on a smooth (frictionless) surface. A 12 N horizontal force is applied to the box for a total of 6.5 seconds. Calculate the work done by the force.

Answer 1

The work done by a constant force F over a distance d (in the same direction as d) is given by

$\\\\W=Fd$

Since the constant force is acting, the distance traveled by the block in time t is

$\\\\d=ut+\frac{1}{2}at^2$

Since it is starting from rest u=0. The acceleration of the block is

$\\\\a=\frac{F}{m}$

Now the distance traveled is

$\\\\W=Fd=F\times \frac{F}{2m}t^2=\frac{F^2t^2}{2m}\\\\W=\frac{(12)^2\times (6.5)^2}{2\times 5.4}\\\\W=563\,J$

Rounding this off to two significant digits we get

$\\\\W=560\,J$

This is the answer.