Question

Velocity of an object of mass 5 kg increases from 3 m/s to 7 m/s on applying a force continuously for 2 s. Find out the force applied. If the duration for which force acts is extended to 5 s, what will be the velocity of the object then? ​

Answer 1

Answer:

The velocity of an object of mass 5 kg increases from 3 m/s to 7 m/s on applying a force continuously for 2 s. The force applied will be equal to 10 N. If the duration for which force acts is extended to 5 s,  the velocity of the object will be $2\ m/s^{2}$

Explanation:

For calculating force (F) applied by the breaks on the tyres, consider Newton's second law of motion Which is stated as,

$F = ma$     ...(1)    

Where,

m -  the mass of the object

a - acceleration

Acceleration can be obtained by calculating the rate of change of velocity with respect to time.

That is,

$a = \frac{v-u}{t}$    ...(2)

first equation become,

$F =m( \frac{v-u}{t})$  ...(3)

where

u -  Initial velocity

v -  Final velocity

t  -  Time

From the question,

$u = 3\ m/s$

$v = 7\ m/s$

$m = 5\ kg$

$t = 2\ s$

Substituting these values into equation (3),

$F =5( \frac{7-3}{2}) =10\ N$ ....(4)

If time extended to 5s,

$t = 5 s$

From equation(1),

$a = \frac{F}{m}$

Substituting the values to the above equation.

we get,

$a = \frac{10}{5} = 2\ m/s^{2}$

   

Answer 2

Answer:

QUESTION:-

Velocity of an object of mass 5 kg increases from 3 m/s to 7 m/s on applying a force continuously for 2 s. Find out the force applied. If the duration for which force acts is extended to 5 s, what will be the velocity of the object then?

ANSWER:-

U = 3m/s

V = 7m/s

t = 2s

m = 5kg

According to second law of motion

F = ma

$= \frac{m(v - u)}{t} \\ \\ = \frac{5(7 - 3)}{2} = 10N$

$Acceleration \ \: : a \: = \frac{ F }{m} = \frac{10}{5} = \\ {2 \:m/s }^{2}$

If we substitute the values in the equation v = u + at , velocity can be calculated when the time for force is extended to 5s

V = 3 + (2×5) = 13m/s.