A shopper in a supermarket pushes a cart with a force of 35.0N directed at an angle of 25 below the horizontal. The force is just sufficient to balance various friction forces, so the cart moves at constant speed.

A Shopper In A Supermarket Pushes A Cart With A Force Of 35N

  1. Calculate the work done by the customer on the cart as she drives down a  long slide.
  2. What is the net work done on the cart? Explain.
  3. The customer goes down the next slide, moving horizontally and keeping the same speed as earlier.  If the friction force doesn’t alter, would the customer’s devoted force be more, less, or unchanged? What do you say regarding the work done on the cart by the customer?

This problem aims to find the work done by the customer on the cart as she slides down the hall. The concepts required for this problem are related to basic physics, which includes work done on a body and friction force.

The concept of work done comes as the dot product of the horizontal component of the force with the direction of the displacement along with the value of the displacement.

\[ F_s = F_x = F\cos \theta \space s \]

The component that is responsible for the movement of the object is $Fcos\theta$, where $\theta$ is the angle between the force $F$ and the displacement vector $s$.

Mathematically, Work done is a scalar quantity and is expressed as:

\[ W = F \times s = (F\cos \theta) \times s \]

Where $W=$ work, $F=$ force exerted.

Expert Answer

Part a:

We are given the following information:

Magnitude of force $F = 35 N$,

The angle at which the force occurs $\theta = 25 $ and,

The displacement $\bigtriangleup s = 50 m$.

To calculate the work done, we are going to use the formula:

\[ W_{customer} = F \times s = (F\cos \theta) \times \bigtriangleup s\]

\[ W = (35.0 N)(50.0 m)\cos 25\]

\[W=1.59\times 10^3\space J\]

Part b:

Since the cart moves at a constant speed,

\[ F_x – f=0 \implies f=+F\cos25 \]

Where $f$ is the work done by friction.

\[ W_f=fx\cos 180^{\circ}\]


\[=-F\cos 35\times x\]


Since $W_{net}=W_s+W_f $

So $W_{net}=0$, as the speed doesn’t change.

Part c:

Since the cart stays at a constant velocity, the force exerted on the cart will be equal to the friction force as it is now completely horizontal to the surface. Thus the net work done on the cart will be equal to the change in kinetic energy being generated due to the change in position. 

\[W_{net}=\bigtriangleup K.E.\]

Since the speed doesn’t change,


We know that the net work done $W_{net}$ is the sum of frictionless work $W_s$ and work under the force of friction $W_f$, so:

\[W_{net}=W_s+W_f \]

\[W_s=-W_f \]

Also, $F_{net}=-f$, which says that the friction is smaller when the customer pushes the cart horizontally.

Numerical Result

Part a: $W=1.59\times 10^3\space J$

Part b: $W_{net}=0$

Part c: $W_s=-W_f$


Find the work done in driving a cart via a distance of $50 m$ against the force of friction of $250N$. Also, comment on the kind of work done.

We are given:

The Force exerted, $F=250N$,

Displacement $S=50m$,

\[ W=F\times S\]


\[=1250\space J\]

Note that the work done here is negative.

Previous Question < > Next Question