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

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The discussion revolves around calculating the total work done by a 94.6 N force acting on a block being slid along a rough surface at a 29-degree angle. The user initially calculated the net force and work done based on that, arriving at approximately 4037.13 J. However, it was pointed out that the question specifically asks for the total work done by the applied force, not the net force. This misunderstanding led to frustration, but the user acknowledged the error and expressed gratitude for the clarification. The key takeaway is the importance of carefully reading the problem statement to determine what is being asked.
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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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