Question

a constant force act on an object of mass 5 kg for a duration of 2 seconds it increase the object lusty from 3 m s-1to 7 m s-1 find the magnitude to the applied force now if the force is applied for a duration of 5 seconds what would be the final velocity of the object

Answer 1

First of all we have to find the acceleration a

We know that a = (v - u)/t

Where v = final velocity
u = initial velocity
and t = time


So substituting the value of v, u and t we get

a =
$\frac{7 - 3}{2} \\ \\ = > \frac{4}{2} \\ = > 2$
Hence acceleration = 2m/s²


Now we know the mass (5kg given) and acceleration we can calculate the force by using Newton's second law i.e

$\boxed{F \:=\:ma}$


Where F is force, m = mass and a= acceleration


So F = 5 × 2

=> F = 10N

Now we know that F = ma

but a = (v - u)/t

=> F = m × (v-u)/t

or, (v - u)/t = F/m

or v - u = Ft/m

=> v = Ft/m + u

So final velocity = Force × time/mass + u

Now substituting the value we get

v = 10 × 5/5 + 3

=> v = 10 + 3

=> v = 13m/s


Hope it helps dear friend ☺️✌️✌️

Answer 2

given mass of object = 5kg

time 1 = 2seconds

u = 3m/s

v1 = 7m/s

v2 = ?

a = ?

We know that a = (v - u)/t

( 7 - 3 ) / 2 = 4 / 2 = 2m/s^2

we get a = 2m/s^2

as the mass = 5kg

by using Newton's second law , F = ma F = 5 × 2 = = 10N.

Now we know that F = ma let us take a as (v - u)/ t substituting we get = F = m × (v-u) / t = (v - u) / t = F / m , v - u = Ft / m= v = Ft / m + u

So final velocity = Force × time/mass + u .

therefore v = 10 × 5 / 5 + 3= v = 10 + 3= v = 13m/s

hope it helps you my friend