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
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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 !
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 !
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