A simple pendulum of length of 5m, with a bob of mass 1 kg, is in simple harmonic motion. As it passes through its mean position, the bob has a speed of $5 \frac{m}{s} .$ The net force on bob at the mean position is

Zero

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2.5 N

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5 N

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25 N

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Solution

The correct option is C
5 N

T- mg = ma_r
$T - m g = \frac{m v^{2}}{r}$
$F_{n e t} = T - m g = \frac{1 \times 5^{2}}{5} = 5 N$

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A simple pendulum of length of 5m, with a bob of mass 1 kg, is in simple harmonic motion. As it passes through its mean position, the bob has a speed of 5ms. The net force on bob at the mean position is

The correct option is C 5 N T- mg = mar T−mg=mv2r Fnet=T−mg=1×525=5N

A simple pendulum of length of 5m, with a bob of mass 1 kg, is in simple harmonic motion. As it passes through its mean position, the bob has a speed of $5 \frac{m}{s} .$ The net force on bob at the mean position is

The correct option is C
5 N

T- mg = ma_r
$T - m g = \frac{m v^{2}}{r}$
$F_{n e t} = T - m g = \frac{1 \times 5^{2}}{5} = 5 N$