Motion of a block on a steep surface

  • #1
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Homework Statement


The block was given an initial velocity up the surface with an angle of 45 degrees to the ground. Calculate the ##\frac{t_1}{t_2}## with ##t_1## being the time it took to get to the highest point up the hill and ##t_2## the time it took to get down. In both cases the frictional force with coefficient m= 0.2 was acting on it.
2. Homework Equations
3. The Attempt at a Solution
My tried solution is uploaded below in form of a picture.
I have made free body diagrams but am stuck at getting the equation that involves time so that i can divide the two times. Any hints?
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Answers and Replies

  • #2
PeroK
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I have made free body diagrams but am stuck at getting the equation that involves time so that i can divide the two times. Any hints?

Think generally: if a body moves a distance ##d## how does the time depend on the acceleration?
 
  • #3
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Think generally: if a body moves a distance ##d## how does the time depend on the acceleration?

##d=v_ot + \frac{1}{2}at^2##
Im still kinda stuck...
 
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PeroK
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##d=v_ot + \frac{1}{2}at^2##
Im still kinda stuck...

The body goes the same distance up the slope as it does down the slope. Down the slope is easy, as ##v_o = 0##.

You can use a trick for motion up the slope: the time taken to move a distance ##d## from an initial velocity to rest at deceleration ##a## is the same as the time to move a distance ##d## from rest at acceleration ##a##. This avoids needing to calculating ##v_0##.

To see this, think about distance as the area under a velocity/time graph.
 
  • #5
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The body goes the same distance up the slope as it does down the slope. Down the slope is easy, as ##v_o = 0##.

You can use a trick for motion up the slope: the time taken to move a distance ##d## from an initial velocity to rest at deceleration ##a## is the same as the time to move a distance ##d## from rest at acceleration ##a##. This avoids needing to calculating ##v_0##.

To see this, think about distance as the area under a velocity/time graph.

I dont understand the trick, so in the new equation for the motion up the slope i would not have ##v_o## but would have some new ##a## while d stays the same right? How would i make that work?
 
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