91Ó°ÊÓ

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

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 \(\left( {d = 20.0\,{\rm{m}}} \right)\).

(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 \( - 700{\rm{ J}}\).

(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 \(0{\rm{ J}}\).

(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 \(700{\rm{ J}}\).

(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 \(38.62{\rm{ N}}\).

(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 \(0{\rm{ J}}\).