Cart Pushing: what happend to s & fk?

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Homework Help Overview

The discussion revolves around a physics problem involving a shopping cart being pushed at a constant velocity. The problem includes the forces acting on the cart, specifically the applied force, frictional force, and gravitational force, as well as the work done by these forces over a specified distance.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are exploring the relationship between force and motion, particularly in the context of constant velocity and net force. Questions arise regarding the role of distance in work calculations and the direction of forces involved.

Discussion Status

There is an active exploration of Newton's laws and their implications for the forces acting on the cart. Some participants are questioning the assumptions made about the forces and their directions, while others are attempting to clarify the relationship between force and work in this scenario.

Contextual Notes

Participants are grappling with the implications of constant velocity, which suggests a balance of forces, and are discussing the relevance of angles in force calculations. There is a noted confusion regarding the application of distance in work calculations and the correct interpretation of the forces involved.

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Cart Pushing: what happened to s & fk??

A person pushes a 16.0 kg shopping cart at a constant velocity for a distance of 22.0 m. She pushes in a direction 29 degrees below the horizontal. A 48.0 N frictional force opposes the motion of the cart. (a) What is the magnitude of the force the shoper exerts? Work done by (b) pushing force, (c) friction force and (d) gravitational force?

(a) I used the formula W=(FcosΘ)s = F(cos29)-Fk = 54.9 N
My reasoning was to take the sum of the two forces (the force minus the frictional force), multiply by cosΘ and solve for F. But where did s (distance) go? I don't understand why it's not used?

(c) Work done by frictional force = -1060 J
I thought the force in this problem was 48 N. And so, I fugured W = 48(cos29)(22) Why is this not correct?
:rolleyes:
 
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Perhaps this can be better explained, but I think of force as an instantaneous quantity. I don't think you should approach part a by thinking about work. What does the fact that the shopping cart is traveling at constant velocity tell you about the net force on the shopping cart?

As for part (c), what direction is the frictional force pointed in?
 
(a) The only thing I can think of is since velocity is constant, there is no acceleration, but I don't see how acceleration effects the net force. I'm sorry, but I'm not seeing something here??

(c)In my diagram, Fk points in the left direction.
 
(a) Think Newton's laws. F = ma. So if a = 0, what does F equal? What does this tell you about Ffrction + Fpush?

(c) Why do you want to multiply by the cosine of the angle?
 

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