Question

a body of mass 10 kg is moving with a velocity of 5 m/s attains of 35m/s in 5 second find the force applied on body to change its velocity
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Answer 1

Correct question:-

A body of mass 10kg moving with a velocity if 5m/s, attains a velocity of

35m/s in 5 seconds. Find the force applied on the body to change it's velocity.

Given:-

→Mass of the body = 10kg

→Initial velocity of the body = 5m/s

→Final velocity of the body = 35m/s

→Time taken = 5s

To find:-

→Magnitude of force applied on the body

Solution:-

By Newton's 2nd law of motion, we know that:-

=> F = ∆p/t

=> F = mv-mu/t. [∵ ∆p = mv-mu]

=>F = m(v-u)/t.

Where:-

•F is force

•m is mass if the body

•v is final velocity of the body

•u is initial velocity of the body

•t is time taken

=> F = 10(35-5)/5

=> F = 10×30/5

=> F = 10×6

=> F = 60N

Thus, the magnitude of force applied on the body to change it's velocity

is 60N.

Answer 2

Explanation:

GIVEN :-

• Initial velocity , u = 5 m/s.

• Final velocity , v = 35 m/s.

• Time , t = 5 seconds.

• Mass of body , m = 10 kg.

TO FIND :-

• The force applied , F

SOLUTION :-

$\\ \underline{ \bigstar \: \textsf{Apply the 1st \: Equation of motion.}}$

$: \implies \displaystyle \sf \:v = u + at$

$: \implies \displaystyle \sf \: 35 \: ms ^{ - 1} = 5ms ^{ - 1} + a \times 5$

$: \implies \displaystyle \sf \:35 \: ms ^{ - 1} - 5 \: ms ^{ - 1} = 5s \times a$

$: \implies \displaystyle \sf \:30 \: ms ^{ - 1} = 5s \times a$

$: \implies \displaystyle \sf \:a = \dfrac{30 \: ms ^{ - 1} }{5}$

$: \implies \underline{ \boxed{\displaystyle \sf \bold{ \:a = 6 \: ms ^{ - 2} }}}$

__________________…

$\underline{ \bigstar \: \textsf{Apply the \: formula \: of \: force .}}$

$: \implies \displaystyle \sf \:Force = mass \times Acceleration$

$: \implies \displaystyle \sf \:Force =10 \: kg \times 6 \: ms^{ - 2}$

$: \implies \displaystyle \sf \:Force =60 \: kg.ms ^{ - 2}$

$: \implies \underline{ \boxed{ \displaystyle \sf \bold{ \:Force =60 \: N}}}$