College Physics by Openstax Chapter 7 Problem 1


How much work does a supermarket checkout attendant do on a can of soup he pushes 0.600 m horizontally with a force of 5.00 N? Express your answer in joules and kilocalories.


Solution:

The work WW that a force FF does on an object is the product of the magnitude FF of the force, times the magnitude dd of the displacement, times the cosine of the angle θ\theta between them. In symbols,

W=FdcosθW=Fd \cos \theta

We are given the following values: F=5.00 NF=5.00\ \text{N}, d=0.600 md=0.600\ \text{m}, and θ=0\theta=0^\circ.

Substitute the given values in the formula for work.

W=FdcosθW=(5.00 N)(0.600 m)cos0W=3.00 NmW=3.00 J  (Answer)\begin{align*} W & = Fd \cos \theta \\ W & = \left( 5.00\ \text{N} \right)\left( 0.600\ \text{m} \right) \cos 0^\circ \\ W & = 3.00\ \text{Nm} \\ W & = 3.00\ \text{J} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The work done is 3.00 Joules. Now, we can convert this in unit of kilocalories knowing that 1 kcal=4186 J1\ \text{kcal} = 4186\ \text{J}.

W=3.00 JW=3.00 J × 1 kcal4186 JW=0.000717 kcalW=7.17×104 kcal  (Answer)\begin{align*} W & = 3.00\ \text{J} \\ W & = 3.00\ \text{J}\ \times \ \frac{1\ \text{kcal}}{4186\ \text{J}} \\ W & = 0.000717\ \text{kcal} \\ W & = 7.17 \times 10 ^{-4} \ \text{kcal} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The work done in kilocalories is about 7.17×1047.17 \times 10 ^{-4}.