Question

A body of mass 2kg moving with a velocity of 10m/s is brought to rest in 10sec . Calculate stopping force required

Answer 1

Answer:

-4N

Explanation:

M= 2kg

V= 10m/s

T= 5sec

AS WE KNOW

V= U + AT

0= 10+ 5a

a = -2m/s

f = ma

= 2 x -2

= -4 (ANSWER)

HOPE IT HELPS YOU.

Answer 2

Answer:

The stopping force required is F = 2 N.

Explanation:

Given,

Mass of the body (m) = 2 Kg

Velocity with which it is moving (u) = 10 m/s

time at which it brought to rest (t) = 10 sec

To find,

The stopping force required

We know that v - u = at

Where, v = final velocity of the body, which is zero in this case.

            u = initial velocity of the body, which is 10 m/s

            a = acceleration of the body with which it had stopped the              

                  vehicle.

            t = time taken by the vehicle to stop, which is 10 sec.

Now putting all into the equation

0 - 10 = (-a)(10)  [As it is decceleration]

 $a =1m/s^2$

Now, from Newton's second law:

F = ma

⇒ F = (2 Kg)($1 m/s^2$)

We get F = 2 N.

Therefore, the stopping force required is F = 2 N.

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