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How do I solve this...A block of mass m is projected across a horizontal surface with an initial speed v?
It slides until it stops due to the friction force between the block and the surface. The same block is now projected across the horizontal surface with an initial speed v/2. When the block has come to rest, how does the distance from the projection point compare to that in the first case?
Ellie McIntyre
Short answer: should be one-fourth of the distance.
Long answer:
You can solve with with the equation
v^2 = u^2 + 2as
We know that the final velocity is zero
s = -u^2/2a
When we sub in u = 1/2 u, it becomes a quarter of the original displacement. So, it will travel a quarter of the distance.
Anonymous
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