How Is Work Calculated When Pulling a Wagon?

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Homework Help Overview

The problem involves calculating work done while pulling a wagon, considering forces acting on the wagon, including friction and the angle of the applied force. The context includes a child and wagon with a specified mass and work done by an adult pulling the wagon at a constant speed.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relationship between work, force, and displacement, questioning the apparent discrepancy between the calculated applied force and the force of friction. There is exploration of the angle at which the force is applied and its effect on the normal force.

Discussion Status

Some participants suggest drawing a free body diagram (FBD) to clarify the forces involved and emphasize the importance of the angle of the applied force. There is recognition of the need to account for vertical components of the force and their impact on the normal force and friction.

Contextual Notes

The problem includes specific values for mass, work done, distance, and the coefficient of friction, which are essential for the calculations but may lead to confusion regarding the relationships between the forces involved.

danrochester
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[SOLVED] Work done on a wagon

Homework Statement


A child and the wagon he/she is riding in has a combined mass of 50 kg, and the adult pulling the wagon does 2.2 x 10^3 J of work pulling the two 60 m at a constant speed. The coefficient of friction for the surfaces in contact is 0.26.
a) draw an FBD for the wagon
b) determine the magnitude of force applied by the parent.
c) determine the angle at which the parent is applying the force

Homework Equations


Well the equations I am using seem to indicate that this is a nonsense question, but here goes:
W=Fd (work equals force times change in displacement)
F(kinetic)=F(normal)mu(K) (force of kinetic friction equals normal force times coefficient of kinetic friction)

The Attempt at a Solution


Since the amount of work is given, I divided it by the displacement and got 36.67 N, which should be the force applied by the adult in the direction of motion.

However, using the formula for kinetic friction I get 127.4 N opposing the motion. If the wagon is moving at constant speed, shouldn't these two be equal? I don't really know where I am going wrong, since from what I understand "work" only applies in the direction of motion, and the friction force only applies in opposition to the direction of motion...needless to say this problem is confusing me, any help would be appreciated.
 
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The trick here is that the force is applied at some angle, which changes the normal force. Draw yourself a FBD showing this force acting at some angle. Then make use of the two facts: (1) The work is given, and (2) The velocity is constant. Use them to solve for the force and the angle.
 
ahhhh that's where you didnt read part c.

Part c asks for the angle which the father is pulling, that means that the parent is applying a force that has a vertical component too, thus lowering the reaction force on the wagon. That means that you have to come up with an equation in part c.the kinetic friction experienced is also dependent on this vertical force that the parent is exerting on the wagon. You ahve to do a subtraction between the weight and the force.
 
I think I've got it now.

Friction force equals applied horizontal force which can be found using equation for work, and also equals given coefficient of friction x the normal force.

Applied vertical force = Gravity force - Normal force

Use pythagorean theorem to find applied force.

For part c) angle = tan^-1 F(applied vertical)/F(applied horizontal)

my answer is 351 N at 84 degrees to the horizontal
 
Looks good to me.
 

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