Chapter 4.1, Problem 4PP

To determine

To calculate: the work done in lifting an object of mass $5\text{ kg}$ through a vertical distance of $2\text{ m}$

Answer to Problem 4PP

$\underset{\_}{98\text{ J}}$ work is done in lifting the object through a vertical distance of $2\text{ m}$ .

Explanation of Solution

Given data:

Mass of the object, $m=5\text{ kg}$

Distance moved vertically, $d=2\text{ m}$

The force needed to lift the object is equal to the gravitational force on the object.

Formula used:

Work Equation:

  $W=Fd$

Where $W$ is work in joules, $F$ is the applied force in newton and $d$ is the distance moved by the object in meters

Calculation:

As per the problem,

The gravitational force exerted on an object is the object’s weight.

Weight = mass (kg) $\times$ gravitational strength (N/kg)

Therefore,

Force needed to lift the object = weight of the object

  $\begin{array}{l}m=5\text{ kg}\\ \text{g = 9}\text{.8 N/kg}\\ F=(5\text{ kg)(9}\text{.8 N/kg)=49 N}\end{array}$

Substitute the values of force, $F=\text{49 N}$ and distance, $d=2\text{ m}$ in the work equation:

  $\begin{array}{l}W=Fd\\ W=(49\text{ N)(2 m)}\\ W=98\text{ J}\end{array}$

Conclusion:

$\underset{\_}{98\text{ J}}$ work is done in lifting the object through a vertical distance of $2\text{ m}$