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Mechanical energy of a sliding block

  • Thread starter Vikingjl11
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  • #1
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Homework Statement


The equation below describes a physical situation:

1/2(1.70kg)(3.30m/s)2 + (1.70kg)(9.8m/s2)(2.35m)sin 30º =

1/2(1.70kg)(1.08m/s)2 + 0.320(1.70kg)(9.8m/s2)(2.35)cos 30º


Which description best fits the equation?

A. A 1.70 kg block slows down while sliding down a frictionless plane inclined at a 30° angle.
B. A 1.70 kg block slows down while sliding down a plane with mk = 0.320, with the plane inclined at a 30° angle.
C. A 1.70 kg block slows down while sliding up a frictionless plane inclined at a 30° angle.
D. A 1.70 kg block slows down while sliding down a plane with mk = 0.320, with the plane inclined at a 30° angle.
E. A 1.70 kg block slides over the top of an inclined plane and then descends on the other side. Both planes, inclined at a 30° angle, have mk = 0.320.


Homework Equations



1/2mv2 = KE mgh(sin 30) = W by gravity


The Attempt at a Solution



The equation shows that the block slows down and there is work done by gravity and friction. I just can't justify any of those answers as being correct. Also, you'll notice answer b and d are identical which is odd. Any help would be greatly appreciated.
 
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Answers and Replies

  • #2
Doc Al
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Homework Statement


The equation below describes a physical situation:

1/2(1.70kg)(3.30m/s)2 + (1.70kg)(9.8m/s2)(2.35m)sin 30º =

1/2(1.70kg)(3.30m/s)2 + 0.320(1.70kg)(9.8m/s2)(2.35)cos 30º
You think those two expression are equal?
 
  • #3
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Sorry, the second velocity should be 1.08m/s. I edited the initial post.
And, yes, the question states that those expressions are equal.
 
  • #4
Doc Al
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Sorry, the second velocity should be 1.08m/s. I edited the initial post.
And, yes, the question states that those expressions are equal.
That's better, but I still don't see how those expressions are equal. (Do the arithmetic and check for yourself.)
 
  • #5
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Yeah, the expressions are only equal when the term for gravity is negative.
 
  • #6
Doc Al
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1/2(1.70kg)(3.30m/s)2 + (1.70kg)(9.8m/s2)(2.35m)sin 30º =

1/2(1.70kg)(1.08m/s)2 + 0.320(1.70kg)(9.8m/s2)(2.35)cos 30º
Would you agree that this equation, as written, cannot be correct? If you meant to write it differently, please correct it.

Are you presenting this problem exactly as it was given? What textbook is it from?
 
  • #7
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Yes, I do agree it could not possibly be correct. The problem is written exactly how it is presented.
It comes from an online homework program. I'm not sure if it is at all related to our textbook (Physics: For Scientist and Engineers - seventh edition, Thomson Brooks/Cole).
 
  • #8
Doc Al
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Yes, I do agree it could not possibly be correct.
Well, there you go. Not much point debating the solution to a badly posed problem.
 
  • #9
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Alright.....well, thanks for your time.
 

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