Question-218013

Question Number 218013 by Tawa11 last updated on 25/Mar/25

Answered by mr W last updated on 26/Mar/25

a)

A P = B P = l = \sqrt{1^{2} + 2 . 4^{2}} = 2.6 m

l_{0} = 2 m

Δ l = l - l_{0} = 2.6 - 2 = 0.6 m

\sin θ = \frac{1}{2.6}

2 T \sin θ = m g

2 T \times \frac{1}{2.6} = m g

\Rightarrow T = 1.3 m g

T = \frac{λ}{l_{0}} Δ l = \frac{637}{2} \times 0.6 = 1.3 m g

\Rightarrow m = \frac{637 \times 0.6}{2 \times 1.3 \times 9.81} \approx 15 k g

b)

l = \sqrt{0 . 7^{2} + 2 . 4^{2}} = 2.5 m

Δ l_{1} = 2.5 - 2 = 0.5 m

\sin θ_{1} = \frac{0.7}{2.5} = 0.28

T_{1} = \frac{λ}{l_{0}} Δ l_{1}

m a = m g - 2 T_{1} \sin θ_{1}

a = g - \frac{2 T_{1} \sin θ_{1}}{m}

= (1 - \frac{Δ l_{1} \sin θ_{1}}{Δ l \sin θ}) g

= (1 - \frac{0.5 \times 0.28 \times 2.6}{0.6 \times 1}) \times 9.81 \approx 3.86 m / s^{2}

Commented by Tawa11 last updated on 26/Mar/25

Thanks sir .

I appreciate .

Commented by Tawa11 last updated on 26/Mar/25

I have seen a mistake in their work sir .

They used m = 12 . 5 kg .

They also proved m = 15 kg .

so, I {don}^{'} t know why they now used m = 12 . 5 kg

to find the acceleration . ?

silly error

Commented by Tawa11 last updated on 26/Mar/25

Commented by mr W last updated on 26/Mar/25

w e e v e n {d i d n}^{'} t n e e d m t o f i n d a :

a = (1 - \frac{Δ l_{1} \sin θ_{1}}{Δ l \sin θ}) g

Commented by Tawa11 last updated on 26/Mar/25

Yes, I observed it in your workings sir .

God bless you always sir .

Facebook Tweet Pin