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abruzzese [7]
2 years ago
13

A 120 -kg crate accelerates toward the positive x-direction . If the magnitude of the force due to friction is 74.4 N , what for

ce must pull on the crate to give it a 2.5 m/s ^ s acceleration ?
Physics
1 answer:
Usimov [2.4K]2 years ago
4 0

Answer:

The applied force to accelerate the crate is 374.4 N

Explanation:

Given;

mass of the crate, m = 120 kg

magnitude of force due to friction, Fk = 74.4N

acceleration of the crate, a = 2.5 m/s²

The net horizontal force on the crate is calculated as;

∑Fx = F - Fk

ma = F - Fk

F = ma + Fk

where;

F is the applied force to accelerate the crate by 2.5 m/s²

F = (120 x 2.5) + (74.4 N)

F = 300 N + 74.4 N

F = 374.4 N

Therefore, the applied force to accelerate the crate is 374.4 N

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Answer:

The tension in the string is quadrupled i.e. increased by a factor of 4.

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The tension in the string is the centripetal force. This force is given by

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It follows that F \propto v^2, provided m and r are constant.

When v is doubled, the new force, F_1, is

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Explanation:

Given

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\Rightarrow a_t=0.6\times 0.4\times 10^{-3}\ m/s^2\\\Rightarrow a_t=2.4\times 10^{-4}\ m/s^2\ \text{or}\ 0.00024\ m/s^2

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