Question

a body has a mass of 50 kg its velocity is brought down from 20 metre per second to 5 metre per second by a resisting force in 5 second the magnitude of resisting force is​

Answer 1

A body has a mass of 50 kg its velocity is brought down from 20 metre per second to 5 metre per second by a resisting force in 5 second the magnitude of resisting force is

Given :-

m = 50 kg

u = 20 m/s

v = 5 m/s

t = 5s

To find :-

The magnitude of resistive force .

Solution :-

First we have to find deaceleration produced by the resistive force in time t = 5s.

Acceleration is given by formula :-

a = v-u/t

Now , put the given value

a = 5-20/5

a = -15/5

a = - 3m/sec^{2}

Here , negative sign show the deaceleration produced by resistive force.

Now,

From Newton 2 nd law of motion,

The magnitude of force is given by :-

f = ma

= 50 ×- 3

= -150n

here, Negative sign is due to the opposite direction of resistive force.

hence,the magnitude of resistive force is 150 N.

Answer 2

Answer:

Explanation:

The answer is -150N

Since F=ma

where a=(v-u)/t

F=50*(5-20)/5

F=50*-15/5

F=10*-15

i.e,F= -150N