Calculating Work Done by Forces on a Moving Object

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SUMMARY

The discussion focuses on calculating the work done by various forces acting on a 16.0 kg shopping cart pushed at a constant velocity of 22.0 m at an angle of 29 degrees below the horizontal, with a frictional force of 48.0 N opposing the motion. The key equations used are W = (F cos θ)d and W = Fd. The net force is zero due to the constant velocity, indicating that the pushing force must equal the frictional force. The calculations for the pushing force and the work done by the pushing, frictional, and gravitational forces are essential for understanding the dynamics involved.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with the concepts of work and energy
  • Knowledge of trigonometric functions in physics
  • Ability to apply the work-energy principle
NEXT STEPS
  • Calculate the magnitude of the pushing force using F = ma
  • Determine the work done by the pushing force using W = (F cos θ)d
  • Analyze the work done by the frictional force and gravitational force
  • Explore the implications of constant velocity on net forces in physics
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Students studying physics, particularly those focusing on mechanics, as well as educators looking for examples of force and work calculations in real-world scenarios.

alex7298
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Homework Statement



A oerson pushes a 16.0kg shopping cart at a constant velocity for 22.0m at an angle of 29 degrees below the horizontal. A 48.0N frictional force opposes the motion of the cart.
a)what is the magnitude of the force the shopper exerts on the cart.
Determine the work done by b) the pushing force, c) the fricitonal force, and d) the gravitational force

Homework Equations



W=(Fcostheta)d
W=Fd


The Attempt at a Solution


well I am not sure where to start because using the equation W=(Fcostheta)d leaves you with two variables.
any help would be appreciated
 
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Well, what does that constant velocity tell you about total force?
 
F=ma
so the net force would be zero, which doesn't make sense because the cart is going forward, so wouldn't there have to be a force greater than the one that opposes it?
 

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