Block on an inclined plane

  • Thread starter ProPM
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  • #1
ProPM
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A block is held stationary on a frictionless inclined plane by means of a string.

The angle of inclination of the plane is 25 degrees, the block has mass 2.6 kg. Calculate the force on the string, you may assume that g = 9.8 m/s2 . Sorry for the poor quality of the attachment but I think you get the idea.

So, its the first time I came across this type of question so I'm not sure whether what I did is correct or not:

a) So the force on the string: Block not moving, thus, equal to parallel component of the weight, hence:

T = sin 25 x (2.6 x 9.8)
T = 10.4 N

b) The string is pulled so that the block is now moving at a constant speed of 0.85 m/s up the inclined plane: Explain why the magnitude of the force on the string is the same now as it was in the first question?

N1 First Law says that if forces are balanced, object must be at rest or moving at a constant velocity, thus, since the weight didn't change, the force on the string must be the same

c) Calculate the power required to move the block at this speed:

P = F x velocity
P = (sin 25) x (9.8 x 2.6) x 0.85
P = 8.8 W

d) State the rate of change of GPE of the block:

So, in 1 second, since moving at 0.85 m/s, block is displaced 0.85 m in one second at an angle of 25. Thus, vertical displacement in one second = sin 25 x 0.85
hence: (mxgxh) = (2.6 x 9.8 x (sin 25 x 0.85)) = 9.15 W

Thanks in advance,
Peter G.
 

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Answers and Replies

  • #2
cgk415
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Looks good to me.
 
  • #3
ProPM
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Thanks, I was afraid that, when calculating the force on the string, thus, finding the parallel component of weight I had not used the right trig law, since that could alter all my other answers. So I did that right?
 

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