Question

Slim Jim, continually maintaining his svelte body, lifts a 70 kg barbell 1.4m above the grounda) How much energy did the barbell have when it was on the ground?b) How much energy does it have after being lifted 1.4m? What kind of energy does it have afterbeing lifted? Where did it come from?c) How much work did Jim do to the lift the object?dIf he lifted it in 1.5s how much power did he use?

Answer 1

Given,

The mass of the barbell, m=70 kg

The height to which Slim Jim lifts the barbell, h=1.4 m

a)

When the barbell was on the ground, it will have zero kinetic energy as it has no velocity. And if assume the height of ground as zero meters, then its potential energy is also zero.

Thus when the barbell was on the ground, its energy was zero joules.

b)

The energy of the barbell when it is at a height of h is given by,

`E=\text{mgh}`

Where g is the acceleration due to gravity.

On substituting the known values,

`\begin{gathered} E=70*9.8*1.4 \\ =960.4\text{ J} \end{gathered}`

Thus the energy that the barbell has after being lifted 1.4 m is 960.4 J

This energy is the energy stored in the barbell due to its position. Thus the energy stored in the barbell is the potential energy.

Slim Jim has to do some work to lift the barbell to the given height. This work done will be stored in the barbell in the form of potential energy. That is, the energy of the barbell is supplied to it from Slim Jim through the work.

c)

All the work done by Jim will be stored in the barbell in the form of potential energy. Thus, the work done by Jim is equal to the potential energy of the barbel.

Therefore, the work done by Jim is 960.4 J

d)

Given,

The time interval, t=1.5 s

The power is given by,

`P=(W)/(t)`

Where W is the work done by Jim.

On substituting the known values,

`\begin{gathered} P=(960.4)/(1.5) \\ =640.27\text{ W} \end{gathered}`

Thus the power used by Jim is 640.27 W