Question

A constant force acting on a body of mass 3.0kg changes it's speed from 2.0m/s to 3.5m/s in 25 sec . The direction of the motion of the body remains unchanged . What is the magnitude and direction of the force ?

Answer 1

We need to find the acceleration first.

We know that,

a = ∆v/∆t or,
a = (v - u)/t

Here, v = 3.5 and u = 2
t = 25

so acceleration = (3.5 - 2)/25

=> 1.5/25

=> 0.06m/s²

Direction remains unchanged which means the force is acting in the same direction as that of the motion that is, positive direction.

From the formula, F = ma,

F = 3 × 0.06

=> F = 0.18N in +ve direction


Hope it helps dear friend ☺️✌️

Answer 2

$\bf{HEY\:BUDDY!!}$

HERE'S THE ANSWER..

______________________________

♠️ We know $\underline{Force\:is\: product \:of\:mass\: and\: Acceleration }$ , i.e

▶️ $\bf{F = m \times a }$

⏺️ To find the force we'll first find Acceleration

♠️ $\underline{Time \:rate \:of \:change\: of\: velocity \:is\: known\:as\: Acceleration \:( a )}$

✔️ Acceleration = ( final velocity - initial velocity ) / time

=> a = ( v - u ) / t

=> a = ( 3.5 - 2.0 ) / 25

=> a = 1.5 / 25

=> a = 0.06 m / s^2

▶️ Now we'll find the force

=> F = m × a

=> F = 3 × 0.06

=> $\boxed{ F\:= 0.18 \: N}$

♠️ $\underline{ As\: the \:force\: is\: positive\: , the \:direction \:will \:be\: the \:same \:in \:which\: object \:was\: moving.}$

HOPE HELPED..

$\bf{JAI \:HIND}$

:)