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2 years ago

8

A box is given a push so that it slides across the floor. How far will it go, given that the coefficient of kinetic friction is

0.15 and the push imparts an initial speed of 3.5m/s?

1 answer:

maksim [4K]2 years ago

8 0

D = v^2 / 2ug

d= 3.5^2 / 0,15 x 9.8 m/s^2

the answer should be around 4.2m

hope this helps

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Answer: -3.49 m/s (to the south)

Explanation:

This problem can be solved by the Conservation of Momentum principle which establishes the initial momentum $p_{i}$ must be equal to the final momentum $p_{f}$ , and taking into account this is aninelastic collision:

Before the collision:

$p_{i}=mV_{o}+MU_{o}$ (1)

After the collision:

$p_{f}=(m+M)V_{f}$ (2)

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$M=1625 kg$ is the mass of the truck

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$V_{f}=\frac{mV_{o}+MU_{o}}{m+M}$ (5)

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