Question

9 A body has a mass of 50 kg. Its velocity is brought
down from 20 m/s to 5 m/s by a resisting force in 5
s. The magnitude of resisting force is
(b) 150 N
(a) 50 N
(c) 750 N
(d) 375 N
​

Answer 1

Answer:-

F = 150 N

Optiona →a

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 :-

$a = \dfrac {v-u}{t}$

Now , put the given value

→$a = \dfrac{5-20}{5}$

→$a = \dfrac{-15}{5}$

→$a = -3 m/s^2$

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 :-

$F = m a$

→$F = 50 \times -3$

→$F = -150 N$

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

hence,the magnitude of resistive force is 150 N.

Answer 2

Answer:

Option A is the correct answer.

F = 150 N

Explanation:

Given Problem:

A body has a mass of 50 kg. Its velocity is brought

down from 20 m/s to 5 m/s by a resisting force in 5 s. The magnitude of resisting force is

(b) 150 N

(a) 50 N

(c) 750 N

(d) 375 N

Solution:

To find :

The magnitude of resistive force.

-----------------

Method:

Given that,

M = 50 kg

u = 20 m/s

v = 5 m/s

t = 5s

Step one is to find deaceleration produced by the resistive force in time

t = 5s.

We know that,

Acceleration formula:

$\bigstar\ a = \dfrac{v-u}{t} \bigstar$

Now,

Plug the values in equation,

$a = \dfrac{5-20}{5}$

$a = \dfrac{-15}{5}$

$a = -3m/s^2$

Here,

This negative sign implies that the deaceleration produced by resistive force.

Now,

We know that Newton 2nd law of motion,

The magnitude of force is:

$F = ma$

$F = 50 \times-3$

$F = -150 N$

Here,

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

Hence,

It implies that the magnitude of resistive force is 150 N.