Question

A force of 200 N acts on a body of mass 20 kg. The force accelerates the body from rest until it attains a velocity of 50 ms−1. Through what distance does the force act?

1 25,000 m
2 125 m
3 4,000 m
4 250 m

Answer 1

Answer:

2) 125 m

Explanation:

$\underline{\bigstar\:\textsf{Given:}}$

• Force (F) = 200 N
• Mass (m) = 20 kg
• Initial velocity (u) = 0 m/s $\:\:\:\:\:$ [As started from rest]
• Final velocity (v) = 50 m/s

$\underline{\bigstar\:\textsf{To\:find:}}$

• Distance (s) = ?

$\underline{\bigstar\:\textsf{Solution:}}$

We know, Force = mass × acceleration. Let's find acceleration first !

$\implies$ 200 = 20 × a

$\implies\:\sf{\cancel{\dfrac{200}{20}}\:=\:a}$

$\sf{\underline{\underline{\therefore\:Acceleration\:=\:10\:m/s^2}}}$

Now, we remember Newton's first equation of motion i.e, v = u + at. By this equation we can find the time taken.

$\implies$ 50 = 0 + 10 × t

$\implies\:\sf{\cancel{\dfrac{50}{10}}\:=\:t}$

$\sf{\underline{\underline{\therefore\:Time\:taken\:=\:5\:seconds.}}}$

Now, we also remember Newton's second equation of motion i.e, $\sf{\underline{s\:=\:ut\:+\:\dfrac{1}{2}at^2}}$

Finally put all the values for distance.

$\longmapsto\:\sf{s\:=\:0\:\times\:5\:+\:\dfrac{1}{\cancel{2}}\:\times\:\cancel{10}\:×\:(5)^2}$

$\longmapsto$ s = 0 + 5 × 25

$\small\longmapsto\underline{\boxed{\sf\purple{s\:=\:125\:m}}}$

Answer 2

☆ QuestioN

A force of 200 N acts on a body of mass 20 kg. The force accelerates the body from rest until it attains a velocity of 50 ms−1. Through what distance does the force act?

___________________________________

☆ AnsWer

125 m is the distance.

given:

• force = 200 N
• mass = 20 kg
• initial velocity = 0 m/s as it was at rest .
• final velocity = 50m/s

to find:

distance = ?

first we need to find the acceleration .

we know that F = m × a ( m is mass and a is acceleration , F is force.)

by using this,

F = m × a

200 = 20 × a

a = 200/20

a = 20/2

a = 10 m/s²

Therefore acceleration is 10 m/s²

Now we have to find the time taken.

we know that by Newton's first equation of motion,

V = u + at

50 = 0 + 10t

50 = 10t

t = 50/10

t = 5 s

therefore time take is 5 s

From second equation of motion,

we know that ,

s = u + ½at²

= 0 + ½ × 10 × 5²

= ½ × 10×25

= 5 × 25

= 125 m

therefore 125 m is the distance