Question

A constant force of friticion of 50 N is acting on a body of mass 200 kg moving intialy with a speed of 15 m/s How long does body take to stop what distance will it cover before covering to rest​

Answer 1

Given :

• A constant force of 50 N is acting on a body of mass 200 kg.
• Speed of a body is 15 m/s. (Initial velocity).

To find :

• How does body take to stop.
• What distance will be cover before covering to rest ?

$\;\large\underbrace{\underline{\sf{Understanding\;the\;Concept}}}$

In the question, We are given A body of mass 200 kg and speed of a body is 15 m/s. We have to find the how does the body will take to stop. Also, What distance will the body covers before covering to rest ?

Solution :

How long does body take to stop = ?

Using the Newton's second law,

$\large \sf F \ = \ ma$

Where,

• F = Force.
• m = mass.
• a = acceleration.

Here,

We need to find the acceleration first.

Put the given values in the above formula.

$\longrightarrow 50 \ = \ 20 \times a$

$\longrightarrow \sf a \ = \ \dfrac {50}{20}$

$\longrightarrow \sf a \ = \ 2.5 \ m/s^2$

$\therefore$ The acceleration of the body is 2.5 m/s².

Now,

Find the time taken by body to stop.

$\sf V \ = \ u \ + \ at$

$\longrightarrow \sf 0 \ = \ 15 \ + \ (-2.5) t$

$\longrightarrow \sf t \ = \ \dfrac {15}{2.5}$

$\large{\red{\boxed{\sf t \ = \ 6 \ seconds}}}$

$\therefore$ The body will take 6 seconds to stop.

Now,

Distance covered by the body = ?

$\sf Distance \ = \ \dfrac {Speed}{time}$

$\longrightarrow \sf Distance \ = \ \dfrac {15}{6}$

$\large{\red{\boxed{\sf Distance \ = \ 2.5 \ meters}}}$

$\therefore$ The distance travelled by the body is 2.5 meters.