Question

displacement of a particle of mass 2kg varies with time as : s=2t square -2t+10)m . find total work done on particle in a time interval 0 to 2 sec

Answer 1

We know that,

ds/dt = Velocity

But, ds/dt = 4t-2 ( Differentiating the Given eqn w.r.t time)

Therefore, V ( at t= 0) = -2 m/s

V ( at t= 2) = 6 m/s

Therefore, Kinetic Energy ( at V= -2) = 0.5*m*v²

= 0.5*2*4 = 4 J

Kinetic Energy ( at V = 6) = 0.5*2*6 = 6 J

Now, Work Done = Difference in K.E

= 6-4 = 2 J


Hope it Helps!

Answer 2

Answer:32J

ds/dt=V

Explanation:on deffentating we get

V=4t-2

Then

at (t=0)

U=-2m/s

At(t=2sec)

V=6m/s

Final kinetic energy=1/2mv^2

From that we get

KE'1=1/2*2*36=>36

Initial kinetic energy=1/2mu^2

From that we get

KE'2=1/2*2*4=>4

Work=KE'1-KE'2=>36-4

WORK=32J

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