Work Done Problem: Solve for Total Work by 94.6 N Force

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SUMMARY

The total work done by a force of 94.6 N acting at an angle of 29 degrees on an 18.7 kg block displaced 70.6 m, with a coefficient of friction of 0.186, requires careful consideration of the force components. The net force (Fnet) was calculated as 57.18 N, leading to an initial work calculation of 4037.13 J. However, the problem specifically asks for the total work done by the applied force, not the net force, which led to confusion in the solution process. Understanding the distinction between total work and net work is crucial for accurate problem-solving.

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Homework Statement



A block is being slid along a rough surface by a force of 94.6 N forming an upward angle with the horizontal of 29 degrees. The mass of the block is 18.7 kg. It is displaced 70.6 m, and the coefficient of friction is .186

what is the total work done by the force.


Homework Equations



Work = (Fnet)(displacement)

friction = coefficient(Fnormal)


The Attempt at a Solution


I drew a free body diagram, and derived these equations:

Fnormal + 94.6sin(29) = mg
Fnet = 94.6cos(29) - friction

From this, I got that Fnormal = 137.3970099 N
friction = 25.55584385 N
Fnet = 57.18318045 N

thus, 57.18318045 N x 70.6 m = 4037.13254 J

Our online program for homework is telling me this is incorrect >.< I'm really not sure what to do.
 
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Hi Jake4! :smile:

Jake4 said:
A block is being slid along a rough surface by a force of 94.6 N forming an upward angle with the horizontal of 29 degrees. The mass of the block is 18.7 kg. It is displaced 70.6 m, and the coefficient of friction is .186

what is the total work done by the force.
…
Fnet = 94.6cos(29) - friction
…
Fnet = 57.18318045 N

thus, 57.18318045 N x 70.6 m = 4037.13254 J

read the question :rolleyes: …

it doesn't ask for the total work done by the net force. :wink:
 
>.<!

wow, got it now... frustrates me more when it's simple misunderstandings like that!

thanks :)
 

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