A child pulls a wagon by the handle along a flat sidewalk.

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A child pulls a trolley with an 80 N force at a 30° angle along a flat sidewalk, moving it 12 m. The mechanical work done by the child is calculated as 148 J using the formula W = F cos(θ) d. The discussion highlights the need to consider the force of friction, which is 34 N, to determine the total or net work done on the trolley. Participants emphasize the importance of correctly applying the angle in calculations and understanding the impact of opposing forces. The total work done on the trolley requires accounting for both the child's work and the work done against friction.
SilentWind12
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Homework Statement


A child pulls a trolley by the handle along a flat sidewalk. She exerts a force of 80.0 N at an angle of 30.0° above the horizontal while she moves the wagon 12 m forward. The force of friction on the trolley is 34 N.

Homework Equations


(a) Calculate the mechanical work done by the child on the trolley.
(b) Calculate the total work done on the trolley

The Attempt at a Solution


W=F(cos)
80 cos(30)[/B]
W=Fcosdeltad
W=80 N(cos30)(12)
=148J

b)Idk how to do it.
 
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SilentWind12 said:
W=80 N(cos30)(12)
=148J
Try that last step again, taking note that it is 30 DEGREES.
For b), what else is doing work on the trolley? How much work? Positive or negative?
 
SilentWind12 said:
(b) Calculate the total work done on the trolley
This could be worded: Calculate the net work done on the trolley.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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