Question

A simple pendulum consist of a 2.0kg mass attached to a spring.it is released from rest at x as shown its speed at the lowest point y is

Answer 1

Answer:

$\huge\mathfrak\red{answer ⤵}$

$mgh = \frac{1}{2} m {v}^{2}$

$v = \sqrt{2gh}$

$v = \sqrt{2 \times 9.8 \times 1.85}$

$v = \sqrt{3r}$

${\huge{\boxed{\mathcal{{v = 6 {ms}^ - 1{} }}}}}$

Explanation:

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Answer 2

⤵answer⤵

mgh = \frac{1}{2} m {v}^{2}mgh=

2

1

mv 2

v = \sqrt{2gh}v= 2gh

v = \sqrt{2 \times 9.8 \times 1.85}v= 2×9.8×1.85

v = \sqrt{3r}v= 3r

{\huge{\boxed{\mathcal{{v = 6 {ms}^ - 1{} }}}}}

v=6ms −1

Explanation:

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