Question

A body of mass 25kg is at rest and a force of 100 newton is applied to it. What will be the velocity after 10 sec?and how much distance will it travel during this time

Answer 1

GiveN :

• Mass (m) = 25 kg
• Force (F) = 100 N
• Time interval (t) = 10 sec
• Initial velocity (u) = 0 m/s

To FinD :

• Distance travelled during this Time

SolutioN :

Use formula for Force :

$\\ \implies \sf{F \: = \: ma} \\ \\ \\ \implies \sf{a \: = \: \dfrac{F}{m}} \\ \\ \\ \implies \sf{a \: = \: \dfrac{100}{25}} \\ \\ \\ \implies \sf{a \: = \: 4} \\ \\ \\ \longrightarrow \underline{\boxed{\sf{Acceleration \: (a) \: = \: 4 \: ms^{-2}}}}$

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Now, use 2nd equation of motion :

$\\ \implies \sf{s \: = \: ut \: + \: \dfrac{1}{2} at^2} \\ \\ \\ \implies \sf{s \: = \: \big( 0 \: \times \: 10 \big) \: + \: \dfrac{1}{2} \: \times \: 4 \: \times \: 10^2} \\ \\ \\ \implies \sf{s \: = \: 0 \: + \: \dfrac{4 \: \times \: 100}{2}} \\ \\ \\ \implies \sf{s \: = \: 4 \: \times \: 50} \\ \\ \\ \implies \sf{s \: = \: 200} \\ \\ \\ \longrightarrow \underline{\boxed{\sf{Distance \: = \: 200 \: m}}}$

Answer 2

Answer:

Explanation:

DATA:

m=25 kg

vi=0 m/s

f=100 N

vf=?

t=10 secs

S=?

SOLUTION:

FROM THE FORMULA OF FORCE WE GET ACCELERATION:

F=ma

100=25 a

100/25=a

a=4 m/s^2

NOW,

FROM FIRST EQUATION OF MOTION:

vf=vi +at

vf=0+(4)(10)

vf=0+40

vf=40 m/s

FINALLY,

FROM THIRD EQUATION OF MOTION:

2as=vf^2-vi^2

2(4)s=(40)^2-(0)^2

8s=1600-0

8s=1600

s=1600/8

S=200 m

RESULT:

                   The velocity of the body is 40 m/s and distance travelled by the body is 200 m.