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3 years ago

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A small block has constant acceleration as it slides down a frictionless incline. The block is released from rest at the top of

the incline, and its speed after it has traveled 6.00 mm to the bottom of the incline is 3.80 m/s.What is the speed of the block when it is 3.00 m from the top of the incline?

1 answer:

ehidna [41]3 years ago

5 0

Answer:

Explanation:

Given

Speed of block at bottom is $v=3.8\ m/s$

Distance traveled $s=6\ m$

initial velocity is zero

using equation of motion

$v^2-u^2=2as$

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

$(3.8)^2-0=2\times a\times 6$

$a=1.203\ m/s^2$

when it is 3 m from top then

$v^2-u^2=2as$

$v^2-0=2\times 1.203\times 3$

$v=2.68\ m/s$

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Answer:

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First we need to define an Frame of reference. Lets put the x axis unit vector $\hat{i}$ pointing east, with the y axis unit vector $\hat{j}$ pointing south, so the positive angle is south of east. For this, we got for the first force:

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