Question

A body of mass 10 kg is moving with a velocity of 1 m/s. Find (i) The magnitude of the force required to stop the body in 10 sec

Answer 1

Answer:

10 m

Explanation:

u = 1 m/s

v = 0 m/s (As the body comes to rest)

t = 10 seconds

So, a = (v - u)/t

So, a = (0 - 1)/10

So, a = -0.1 m/s^2

(minus sign indicates that the object is decelerating i.e. its velocity is decreasing)

• Mass of the body (m) = 10 kg

Force (F) = m × a

So, F = 10 × (-0.1)

So, F = -1 N

(Here, minus sign indicates that the force is applied in the opposite direction of the motion of the body)

Hence, force applied is 1 newton.

• According to the second kinematical equation of motion,

v^2 = u^2 + 2as

So, s = (v^2 - u^2) / 2a

So, s = (0 - 1)/(-0.1)

So, s = (-1)/(-0.1)

So, s = 10 m

Therefore the body travels 10 m before coming to rest.

Answer 2

Answer:

10 m

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