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sergij07 [2.7K]

3 years ago

13

A force in the negative x-direction is applied for 27 ms to a 0.4 kg mass initially moving at +14 m/s in the x-direction. The fo

rce varies in magnitude and delivers an impulse with a magnitude of 32.4 N-s. What is the mass's velocity in the x-direction? Give the vector component in m/s.

1 answer:

anyanavicka [17]3 years ago

7 0

It is given that,

Mass of the object, m = 0.5 kg

A force is applied for 27 ms.

Initial speed of the object, u = +14 m/s

Impulse of the force, J = 32.4 N-s (in negative direction)

Let v is the velocity of mass in x direction. We know that the impulse of an object is equal to the change in momentum. It is given by :

$J=m(v-u)$

$-32.4=0.4\times (v-14)$

v = -67 m/s

So, the velocity of the mass is 67 m/s and it is in negative x direction.

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Answer:

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Explanation:

Given:

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∴mass, $\Rightarrow m= 45.8163 kg$

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(a)

Now,

The force to be compensated for being on the verge of snapping:

(T-w) = 62 N

Therefore, we need to produce and acceleration equivalent to the above force.

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(b)

From the equation of motion ,we have:

$v^{2} =u^{2} +2a.s$ ....................(2)

where:

u= initial velocity= 0 (here, starting from rest)

v= final velocity = ?

$a= 1.4 m.s^{-2}$

s= displacement =h =6.1 m

Now, putting the values in eq. (2)

$v^2= 0^2 + 2\times 1.4\times 6.1$

$v\approx 4.133 m.s^{-1}$ is the velocity with which the body will hit the ground in the given conditions.

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Given that,

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$I=117.45\ kg-m^2$

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