A simple pendulum is released from a as shown. If m and l represents the mass of the bob and length of the pendulum respectively, the gain in kinetic energy at b is
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at point A, potential energy = mgl
at point B, potential energy = mg(l - lcos30°)
we know for simple pendulum, gain in kinetic energy = -loss in potential energy
= -(potential energy at B - potential energy at A)
= -[mg(l - lcos30°) - mgl ]
= mglcos30°
= √3mgl/2
hence, gain in kinetic energy at B = √3mgl/2
at point B, potential energy = mg(l - lcos30°)
we know for simple pendulum, gain in kinetic energy = -loss in potential energy
= -(potential energy at B - potential energy at A)
= -[mg(l - lcos30°) - mgl ]
= mglcos30°
= √3mgl/2
hence, gain in kinetic energy at B = √3mgl/2
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This is the answer and full complete method of this sum. Hope this will help you . Physics is a great subject . All the best.
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