Question

a body of mass 2 kg is moving along north east direction with a speed of √2 m/s. A force of 0.2 N is applied on the body due west for 10 sec. the final velocity of the body is​

Answer 1

a body of mass 2kg is moving along North-east direction with speed of √2 m/s.

so, initial velocity vector of body, $\vec{u}=\sqrt{2}(cos45^{\circ}\hat{i}+sin45^{\circ}\hat{j})$

or, $\vec{u}=\hat{i}+\hat{j}$

given, force acting on body , $F=-0.2\hat{i}$ for 10 sec

from Newton's 2nd law, F = ma

so, $\vec{a}$= F/m = $-\frac{0.2}{2}\hat{i}=-0.1\hat{i}m/s^2$

using formula,

$\vec{v}=\vec{u}+\vec{a}t$

= (i + j) + (-0.1i) × 10

= j

hence, final velocity of body is 1 m/s along north direction.