Question

A particle of mass 2 kg moves on a horizontal plane under the action of force F=(2ˆi+2ˆj)N. The body is displaced from (0,0) to (1m,1m) If the initial speed of the particle is 2m/s find its final speed

Answer 1

Force = mass * acceleration

So,
acceleration = 1/2*(2i+2j) = i+j
So,
Magntiude of acceleration = √2 m/s^2

Now,
Since that particle is taken from (0,0) to (1,1)
The displacement is √2 metres.

So, from the third equation of kinematics,
v^2 - u^2 = 2as
V^2 - (2)^2 = 2*√2*√2
v^2 = 4 + 4
So,
final velocity (v) = 2√2 m/s

Hope this helps you !