Question

A shopper pushes a grocery cart 20.0 m at constant speed on level ground, against a 35.0 N frictional force. He pushes in a direction 25.0º below the horizontal.

(a) What is the work done on the cart by friction?

(b) What is the work done on the cart by the gravitational force?

(c) What is the work done on the cart by the shopper?

(d) Find the force the shopper exerts, using energy considerations.

(e) What is the total work done on the cart?

Short answer

(a) The work done by friction is\( - 700{\rm{ J}}\).

(b) Work done by the gravitational force is\(0{\rm{ J}}\).

(c) The work done by the shopper is\(700{\rm{ J}}\).

(d) The force that a shopper exerts is\(38.62{\rm{ N}}\).

(e) The total work done is \(0{\rm{ J}}\).

Step-by-step solution

Definition of Concept

Work: Work is the product of the component of force in the displacement direction and the magnitude of the displacement.

Free body diagram of the cart

Free body diagram of the cart

Here, \(F\) is the force exerted by the shopper on cart, \(\theta \)is the angle at which the shopper applies force on the cart \(\left( {\theta = 25^\circ } \right)\), f is the friction force between the floor and the cart \(\left( {f = 35\,{\rm{N}}} \right)\), d is the displacement of the cart \(\left( {d = 20.0\,{\rm{m}}} \right)\).

Calculate work done by the frictional force

(a)

The work done by the frictional force is,

\({W_{fric}} = fd\cos \varphi \)

Here, \(\varphi \) is the angle between the frictional force and the displacement of the car \(\left( {\varphi = 180^\circ } \right)\).

Putting all known values,

\(\begin{aligned}{}{W_{fric}} &= \left( {35.0{\rm{ N}}} \right) \times \left( {20.0{\rm{ m}}} \right) \times \cos \left( {180^\circ } \right)\\ &= - 700{\rm{ J}}\end{aligned}\)

Therefore, the required work is done by the frictional force is \( - 700{\rm{ J}}\).

Calculate work done on the cart by the gravitational force

(b)

The work done by the gravitational force is given as,

\({W_g} = mg\left( {\Delta h} \right)\)

Here,\(m\)is the mass of the cart,\(g\)is the acceleration due to gravity and\(\Delta h\)is the change in height.

Since, the cart moves on a level ground, hence there will no change in height. As a result, the gravitational force's work on the cart is,

\({W_g} = 0{\rm{ J}}\)

Therefore, the required work is done by the gravitational force is \(0{\rm{ J}}\).

Calculate work done on the cart by the shopper

(c)

Because the work done by friction is\( - 700{\rm{ J}}\)and the block moves at a constant velocity, the work done by the shopper on the cart is equal to the work done by friction but in the opposite direction.

\(W = - {W_{fric}}\)

Putting all known values,

\(\begin{aligned}{}W &= - \left( { - 700{\rm{ J}}} \right)\\ &= 700{\rm{ J}}\end{aligned}\)

Therefore, the required work is done by shopper is \(700{\rm{ J}}\).

Calculate the force the shopper exerts

(d)

The shopper's work is as follows:

\(W = Fd\cos \theta \)

Here, \(F\) is the force exerted by the shopper on cart, \(\theta \) is the angle at which shopper applies force on the cart \(\left( {\theta = 25.0^\circ } \right)\).

The shopper's force is denoted by the expression,

\(F = \frac{W}{{d\cos \theta }}\)

Putting all known values,

\(\begin{aligned}{}F &= \frac{{700{\rm{ J}}}}{{\left( {20.0{\rm{ m}}} \right) \times \cos \left( {25^\circ } \right)}}\\ &= 38.62{\rm{ N}}\end{aligned}\)

Therefore, the required force exert by the shopper on the cart is \(38.62{\rm{ N}}\).

Find the total work done on the cart

(e)

The total work done on the cart is,

\({W_{tot}} = {W_{fric}} + W\)

Putting all known values,

\(\begin{aligned}{}{W_{tot}} &= \left( { - 700{\rm{ J}}} \right) + \left( {700{\rm{ J}}} \right)\\ &= 0{\rm{ J}}\end{aligned}\)

Therefore, the required total work done on cart is \(0{\rm{ J}}\).