Determine the work done by the applied force

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Homework Help Overview

The discussion revolves around a physics problem involving work done by various forces on a block being pushed across a frictionless surface. The block has a mass of 2.50 kg and is subjected to a constant force of 12.0 N directed at an angle of 25.0° below the horizontal.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of work done by the applied force, normal force, gravitational force, and net force. There is confusion regarding the angles involved and the direction of the applied force. Some participants question the assumptions about the forces acting on the block and the reasoning behind the calculations for work done.

Discussion Status

Some participants have confirmed the correctness of the original poster's answers for parts a, b, and c, while others are seeking clarification on the direction of the applied force and its implications for the calculations. There is an ongoing exploration of the reasoning behind the work done by different forces, with no explicit consensus reached yet.

Contextual Notes

Participants note that the problem does not provide additional information beyond the stated conditions, leading to questions about the assumptions made regarding the forces and their directions.

BunDa4Th
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Okay I am confuse on how i came to this conclusion and how i came to that answer.

A block of mass 2.50 kg is pushed 2.80 m along a frictionless horizontal table by a constant 12.0 N force directed 25.0° below the horizontal.

(a) Determine the work done by the applied force.
30.45J
(b) Determine the work done by the normal force exerted by the table.
0 J
(c) Determine the work done by the force of gravity.
0J
(d) Determine the work done by the net force on the block.
30.45J

I am confuse on b, c, d. I know that to determine the applied force i would use W = (Fcostheta)(deltaX).

For b i would use W = (mgcos90)deltaX is that correct? because since the block is on a table which is flat and normal force is pointing up so from table to normal force it makes a 90 degree angle.

For c I did a guess since I was not sure but i thought that it would be zero since there is no incline and gravity is not acting on this block because its not falling.

D i came up with Wnet = W_F + W_f + W_N + W_mg
W_F = 30.45
W_f = 0 (no friction so it is zero)
W_N = 0 (not sure why)
W_mg = 0 (not sure why)

sorry its a bit long but can anyone explain this to me.
 
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can u clarify the direction of the force on the object? what u mean by a constant force of 12 N directed below the horizontal? So the force is acting 25 degrees upwards on the block or downwards?
 
that is all the info was given and that's the exact question given.

I think the force is acting 25 degrees downwards from the block.
 
For a, b and c, the answers u provided are right.. if the force on the block is really acting downwards from the block. for a) All u have to do is to take the horizontal component of the force on the block and multiply it by the distance traveled which i think u got it. For both b and c, the WD equals zero because there is no distance traveled in the direction of the force. for d, all u got to do is to sum up the WD on the block since they are scalars which will actually give u the wd done by moving the block, 30.45.
 
Thanks, that make much more sense now.
 

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