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danrochester
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[SOLVED] Work done on a wagon
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
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)
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.
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.