Question

A box of mass 5.0 kg is accelerated from rest by a force across a floor at a rate of 2.0 m/s2 for 7.0 s. find the net work done on the box.

Answer 1

W=Fd. Force is not given so we solve for it. F=ma, m=5kg, a=2m/s^2, F=10N. Distance is not given so we solve for it, x=.5a(t^2)=.5(2)(7x7)=49m. F=10N, d=49m, W=490J.

Answer 2

Answer:

Work done on the box is 490 joules.

Explanation:

It is given that,

Mass of the box, m = 5 kg

Initial speed of the box, u = 0

Acceleration of the box, $a=2\ m/s^2$

Time taken, t = 7 s

The force is calculate using the product of mass and acceleration. It is given by :

F = ma

$F=5\ kg\times 2\ m/s^2=10\ N$

Let d is the distance covered by the box. Using the equation of motion as :

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

$d=\dfrac{1}{2}\times 2\ m/s^2\times (7\ s)^2$

d = 49 m

The product of force and distance is equal to the work done on the box. It is given by :

$W=F\times d$

$W=10\ N\times 49\ m$

W = 490 joules

So, the work done on the box is 490 Joules. Hence, this is the required solution.