A car of mass m is moving along a circular track of
radius r with a speed which increases linearly with time
tas v = kt where k is a constant. The instantaneous
power delivered by agent applying the force is
Answers
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ANSWER ⏬
✏️ Since, the car is moving in circular path and its velocity's magnitude is changing and obviously the direction of velocity will change, this means that there will be two component of the acceleration, one parallel to the path and other Perpendicular to the path. I means there will be Centripetal acceleration and tangential acceleration.
Now, the Net acceleration produced by them will not be along the Centre.
The magnitude of the Frictional force must be less than the mv²/r so that car can remains in the path.
⇒ μmg ≥ mv²/r
⇒ μg ≥ a
⇒ μ ≥ a/g.
This tells us that coefficient of the Friction cannot be less than a/g.
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MARK ANSWER AS BRAINLIST
Answer:mk^2t
Explanation:
Given,
V=kt
Therefore a=v/t
=Kt/t
=K
Now
Power= Fv
=mac
=mk.kt
=mk^2t