Why is the magnitude of its acceleration less on the descent

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SUMMARY

The discussion addresses the difference in acceleration magnitude of a block sliding up and down a ramp, emphasizing the role of friction. When ascending, friction acts downward, opposing the block's motion, while during descent, friction acts upward, also opposing motion. This results in a greater net force acting on the block during ascent compared to descent. The analysis can be further supported by applying the principles of Conservation of Energy and resolving the forces acting on the block.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Basic knowledge of frictional forces
  • Familiarity with gravitational potential energy (GPE)
  • Ability to resolve forces in physics problems
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  • Study the effects of friction on motion in inclined planes
  • Learn about Conservation of Energy in mechanical systems
  • Explore the mathematical analysis of forces using free-body diagrams
  • Investigate the impact of different surface materials on friction coefficients
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katana
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This is sort of a concept problem.
It says: A block is given a push so that it slides up a ramp. AFter the block reaches its highest opint, it slides back down. Why is the magnitude of its acceleration less on the descent than on the ascent?

I know it as to do with friction but I don't see how the friction forces are different on the ascent and the descent.
 
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Here's a hint: Friction always acts to oppose slipping between surfaces. So... when the block slides up the ramp, which way is it slipping with respect to the ramp surface? So which way must friction act? Do the same analysis for the block sliding down the ramp. Then compare total force on the block in both cases.
 
You could also try to analyse it in terms of Conservation of Energy if you want. When you initially give the box a push, the box will use up its kinetic energy to gain GPE and do work against friction. Afterwards, the GPE will be used to do work against friction again and give the box kinetic energy to reach the bottom.

Another way of tackling this is to resolve the forces acting on the box, which you will find:
Going up: mg sin @ + friction (both downwards)
Going down: mg sin @ (downwards) - friction (upwards)
 

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