Two-or-three-dimensional-motion, Mechanical Engineering

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Net force acting on particle of mass 2 kg is 10N in north direction.At t=o, particle was moving eastwards with 10m/s. Find displacement and velocity of particle after 2 seconds.

Ans)
given force =F=10N

mass M=2kg

velocity u=10m/s

then acceleration a=F/M=10/2=5m/s^2

then displacement=s=ut+1/2at^2

here u=initial velocity

t=time

a=acceleration

then we get the displacement is=30m

2.velocity=u+at

then we get velocity as 20m/s.

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3/8/2013 7:09:55 AM Initial velocity is 10 m/s in x-direction. Now a force of 10N acts in y-direction So , using F = ma a = f/m = 10/2 = 5m/s²in y-direction After 2 seconds velocity in y-direction would become 10 m/s Now we can take the resultant of the two velocities i.e 10 m/s in x-direction & 10m/s in y-direction Therefore Resultant velocity = (10²+10²)^1/2 = (200)^1/2 and 45⁰North of East Displacement = velocity x time = (200)^1/2x 2 = 20x(2)^1/2m 45⁰ north of east

3/8/2013 7:09:55 AM

Initial velocity is 10 m/s in x-direction.

Now a force of 10N acts in y-direction

So , using F = ma

a = f/m = 10/2 = 5m/s²in y-direction

After 2 seconds velocity in y-direction would become 10 m/s

Now we can take the resultant of the two velocities i.e 10 m/s in x-direction & 10m/s in y-direction

Therefore Resultant velocity = (10²+10²)^1/2

= (200)^1/2 and 45⁰North of East

Displacement = velocity x time

= (200)^1/2x 2

= 20x(2)^1/2m 45⁰ north of east

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