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Force Pushing a Block on a Rough Surface

  • Thread starter mecablaze
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Homework Statement



A block of mass m slides on a rough horizontal surface from point A to point B. A force (magnitude P = 2.0 N) acts on the block between A and B. P acts on the block at 40 degrees above the horizontal. Points A and B are 1.5 meters apart. If the kinetic energies of the block at A and B are 5.0 J and 4.0 J, respectively.

a) How much work is done on the block by the force P as it moves from A to B
b) How much work is done on the block by the force of friction as the block moves from A to B
c) If the coefficient of kinetic friction is 0.7, what is the mass m?

Homework Equations



kinetic energy = 0.5 * m * v2
horizontal force on the block = cos(40 degrees) * 2.0 N

The Attempt at a Solution



Well, I reasoned that the net change in kinetic energy is -1.0 J from A to B. We divide that by the change in distance so the net force acting on the block from A to B is -1.0/1.5 Newtons. Part of that is due to friction ([tex]\mu[/tex]*(mg + vertical component of P)) and part of it is due to the horizontal component of P. Obviously, since the kinetic energy is decreasing, the force due to friction is greater than the horizontal component of P.

The net force on the block is the sum of the two horizontal force vectors. I just don't know how to split them up and get parts a) and b)

edit: formatting
 
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Answers and Replies

  • #2
PhanthomJay
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Homework Statement



A block of mass m slides on a rough horizontal surface from point A to point B. A force (magnitude P = 2.0 N) acts on the block between A and B. P acts on the block at 40 degrees above the horizontal. Points A and B are 1.5 meters apart. If the kinetic energies of the block at A and B are 5.0 J and 4.0 J, respectively.

a) How much work is done on the block by the force P as it moves from A to B
b) How much work is done on the block by the force of friction as the block moves from A to B
c) If the coefficient of kinetic friction is 0.7, what is the mass m?

Homework Equations



...
horizontal force on the block = cos(40 degrees) * 2.0 N
from this, you can calculate the work done by P in part (a).

The Attempt at a Solution



Well, I reasoned that the net change in kinetic energy is -1.0 J from A to B. We divide that by the change in distance so the net force acting on the block from A to B is -1.0/1.5 Newtons. Part of that is due to friction ([tex]\mu[/tex]*(mg + vertical component of P))
check your plus/minus sign in your equation for friction force
and part of it is due to the horizontal component of P. Obviously, since the kinetic energy is decreasing, the force due to friction is greater than the horizontal component of P.
yes
The net force on the block is the sum of the two horizontal force vectors. I just don't know how to split them up and get parts a) and b)
You are going about the problem out of sequence. You can calculate the work done by P directly, for part a, then use the work energy theorem to calculate the work done by friction, for part b. Then correct your friction force equation to solve for m in part c.
 

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