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

The product of the magnitude of the force applied to the body, the displacement of the body, and the cosine of the angle between the force vector and the displacement vector is the work done.Mathematically,

\(W = Fd\cos \theta \)

Here, \(F\) is the magnitude of the force, \(d\) is the magnitude of the displacement and \(\theta \) is the angle.

Putting\(5.00{\rm{ N}}\)for\(F\),\(0.600{\rm{ m}}\)for\(d\)and\(0^\circ \)for\(\theta \)in equation (1.1),

\(\begin{aligned}{}W &= \left( {5.00{\rm{ N}}} \right) \times \left( {0.600{\rm{ m}}} \right) \times \cos \left( {0^\circ } \right)\\ &= 3.00{\rm{ J}}\end{aligned}\)

Hence, the work done by the supermarket checkout attendant is \(3.00{\rm{ J}}\).

We know that,\(1{\rm{ kcal}} = 4186{\rm{ J}}\). Therefore, the work done is,

\(\begin{aligned}{}W &= \left( {3.00{\rm{ J}}} \right) \times \left( {\frac{{1{\rm{ kcal}}}}{{4186{\rm{ J}}}}} \right)\\ &= 7.167 \times {10^{ - 4}}{\rm{ kcal}}\end{aligned}\)

Hence, the work done by the supermarket checkout attendant is \(7.167 \times {10^{ - 4}}{\rm{ kcal}}\).