Question

a body of mass 6 kg is moving with a velocity of 2 metre per second calculate the force required to stop it in 5 seconds how far will it travel before coming to rest​

Answer 1

Answer:

2.4 N and 0.83 m

Explanation:

Acceleration of body = Change in velocity / Time taken

= (0 m/s - 2 m/s) / (5 s)

= -0.4 m/s²

Force = ma

= 6 kg × -0.4 m/s²

= -2.4 N

∴ A force of 2.4 N should be applied in the opposite direction of motion of body.

Stopping distance = (v² - u²)/(2a)

= (0² - 2²) / (2 × -2.4)

= (4/4.8) m

= (5/6) m

= 0.83 m

Answer 2

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It is given that :-

• Mass of body = 6kg
• Initial velocity = 0
• Final velocity = 2m/s
• Time = ?
• Force required to stop = ?

Acceleration of this body = v /t = 2/5 = 0.4

Now the force required to stop body will be :

$f = ma \\ \\ f = 6 \times 0.4 = 2.4n$

Therefore 2.4N force is required to stop body.

Now according to third equation of motion, square of final velocity of body is equals to sum of its initial velocity and double of product of acceleration and distance. Represented by :-

${v}^{2} = {u}^{2} + 2as$

$s = \frac{ {v}^{2} - {u}^{2} }{2a}$

After inputting the values we get :-

$s = \frac{ {2}^{2} - {0}^{2} }{0.4 \times 2}$

$s = \frac{4}{0.8} = 5$

Therefore it travels 5m while getting stopped.

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BY SCIVIBHANSHU

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