How Do You Calculate the Force Exerted by a Shopper on a Grocery Cart?

[crysland]

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.

Find the force the shopper exerts, using energy considerations.

I got stuck on this because they don't give the mass of the cart.

 

[haruspex]

How would you use the mass of the cart? Suppose it were given, what equations would you apply?
(I find it strange that they ask you to use energy considerations though.)

 

[crysland]

I need to find the x-component of the weight so I need to know the mass.

 

[haruspex]

I need to find the x-component of the weight so I need to know the mass.

What x component? Gravity acts vertically and this is a horizontal surface. It's the force applied by the pusher that is not horizontal.

 

[Nickie]

I would approach this problem by first defining the variables and assumptions. The given information states that the grocery cart is pushed at a constant speed of 20.0 m on level ground, with a frictional force of 35.0 N acting against it. The direction of the push is also given as 25.0º below the horizontal.

To solve this problem using energy considerations, we can use the principle of conservation of energy, which states that the total energy of a system remains constant. In this case, the total energy of the system is the kinetic energy of the cart and the work done by the shopper. We can express this as:

KE + W = constant

Since the cart is moving at a constant speed, its kinetic energy remains constant. Therefore, the work done by the shopper must be equal to the work done by the frictional force. We can express this as:

Wshopper = Wfriction

Now, we can use trigonometric functions to determine the components of the force exerted by the shopper. The force exerted by the shopper can be broken down into two components: one in the direction of motion (horizontal) and one perpendicular to the direction of motion (vertical). We can express these components as:

Fhorizontal = Fcos(25.0º)
Fvertical = Fsin(25.0º)

Now, we can substitute these values into our equation for work:

Wshopper = Fhorizontal * distance = Fcos(25.0º) * 20.0 m

And for the frictional force:

Wfriction = Ffriction * distance = 35.0 N * 20.0 m

Since we know that Wshopper = Wfriction, we can set these two equations equal to each other and solve for the force exerted by the shopper:

Fcos(25.0º) * 20.0 m = 35.0 N * 20.0 m
Fcos(25.0º) = 35.0 N
F = 35.0 N / cos(25.0º)
F = 38.0 N

Therefore, the force exerted by the shopper is approximately 38.0 N. This value may change if different assumptions are made, such as the mass of the cart or the angle of the push. It is important to always clearly define and consider all variables in a problem to