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

Work done on the box is 490 joules.

Step-by-step 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* 2\ m/s^2=10\ N`

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

`d=ut+(1)/(2)at^2`

`d=(1)/(2)* 2\ m/s^2* (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* d`

`W=10\ N* 49\ m`

W = 490 joules

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

Answer 2

The net work done on the box weighing 5.0 kg and accelerated from rest by a force across a floor is 490 Nm.

Given the following data:

• Initial velocity = 0 m/s (since the box starts from rest).

• Mass of box = 5 kg

• Acceleration = 2
  `m/s^2`

• Time = 7 seconds

To find the net work done on the box:

First of all, we would determine the force acting on the box:

`Force = Mass` ×
`acceleration`

`Force = 5` ×
`2`

Force = 10 Newton

Next, we would use the second equation of motion to determine the distance traveled by the box:

`S = ut+ (1)/(2)at^2`

Where:

• S is the displacement or distance traveled.

• u is the initial velocity.

• a is the acceleration.

• t is the time measured in seconds.

Substituting the given parameters into the formula, we have;

`S = 0(7)+ (1)/(2)(2)(7^2)\\\\S = 0 + 49`

Distance, S = 49 meters.

Now, we can determine the net work done on the box:

`Work\; done = Force` ×
`distance`

`Work\; done = 10` ×
`49`

Work done = 490 Nm

Therefore, the net work done on the box is 490 Nm.