A 0.2 kg bob on a 1 m string is released from 6 0 ∘ . Find the tension at the lowest point, in N ( g = 10 ).

v 2 = 2 g l ( 1 − cos 6 0 ∘ ) = 10 . T = m g + m v 2 / l = 2 + 0.2 × 10 = 4 N.

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PhysicsSHMOscillations and SHMEasy2 min

PhysicsEasyinteger

A 0.2 kg bob on a 1 m string is released from $6 0^{\circ}$ . Find the tension at the lowest point, in N ( $g = 10$ ).

Answer:

Solution:

$v^{2} = 2 g l (1 - cos 6 0^{\circ}) = 10$ . $T = m g + m v^{2} / l = 2 + 0.2 \times 10 = 4$ N.

Subject:PhysicsTopic:SHMSubtopic:Oscillations and SHM

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