Chapter 9, Problem 35A

Calculate the work done in lifting a 500-N barbell 2.2 m above the floor. (What is the potential energy of the barbell when it is lifted to this height?) $\text{Power}=\frac{\text{work}\text{done}}{\text{time}\text{interval}}$

To determine

Work done and potential energy.

Answer to Problem 35A

Work done and potential energy will be equal to 1100 J.

Explanation of Solution

Given:

Force is given, $F=500\text{ N}$

Barbell is lifted to a height of, $h=2.2\text{ m}$

Formula used:

According to work energy theorem work done is equal to change in potential energy.

  $W=PE$

Where, W is work done and PE is potential energy.

Potential energy is given by,

  $PE=mgh$

Where, m is mass, g is acceleration and h is height.

Calculation:

Force is equal to,

  $F=mg=500\text{ N}$

Now, substituting the value of mg and h and solve.

  $\begin{array}{l}PE=500\times 2.2\\ =1100\text{ J}\end{array}$

Now, according to work energy theorem,

  $W=PE=1100\text{ J}$

Conclusion:

Therefore, work done and potential energy will be equal to 1100 J.