Question

A person pushes a 16.0 kg shopping cart at a constant velocity for a distance motion She pushes in a direction 29.0 degree below the horizontal. A 48 N frictional force opposes the motion of the cart, What is the magnitude of the force that the shopper exerts? Determine the work done by (b) the pushing force, (c) the frictional force, and (d) the gravitational force.

Help me please

Answer 1

We have the following data to consider:

• Mass: m=16kg.
• Angle: -29º.
• d=22m
• The acceleration is equal to 0 because there is no change of speed.
• The frictional force: 48N.

Making a free-body diagram on the object, and considering that the acceleration is different to 0, we have:

$\sum F_{x}\rightarrow -\vec{F}_{f}=ma\cos (29^{o})$

$\sum F_{y}\rightarrow W-N=ma\sin (29^{o})$

As there is no change of speed, the acceleration is 0, then we have:

$\sum F_{x}\rightarrow \vec{F}_{f}=0$

$\sum F_{y}\rightarrow mg= N$

The definition of Work done by a force is equal to:

$W= \vec{F}\cdot \vec{d}= Fd\cos \Theta$

This states that if the force and the displacement are perpendicular between each other, then the work is equal to 0. As the movement is in the horizontal direction, the normal force and the gravitational force makes no work. The friction force is equal to 0 (considering that the acceleration is 0 for the pushing force), then, it makes no work neither.

You have to remain that the force will be different to 0 as long the acceleration is different to 0, this is, if there is a variation in the speed:

$a= \frac{V_{f}-V_{i}}{t}$

If there is something more that I could help you leave a comment below, regards.