Question

Q.A block of mass 2 kg is being brought down by a
string. If the block acquires a speed 1 m/s in dropping down
25 cm, find the work done by the string in the process.​

Answer 1

Answer:

hope it helps...

Good luck

Answer 2

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✤ Required Answer:

✒ GiveN:

• A mass of a block = 2 kg
• Speed acquire by it = 1m/s
• Displacement of block is = 25cm

✒ To FinD:

Work done by string ?

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✤ How to solve?

A mass of a block is 2 kg which is attached by a string and being bought down . It acquires a speed of 1m/s when the displacement of a block is 25 cm from its initial condition And we need to find the work done by the string in the process .

First to understand the question more nicely let's draw the figure . [ See the attachment - 1 ]

Now we can solve it by using the formula of Newton 2nd law and then by it we need to find the tension and acceleration .

At last in order to find the value of work done we need to apply the formula given below.

☕ Work done :-

$\dag\:\underline{\boxed{\bf{\orange{W=\overrightarrow{F}\cdot\overrightarrow{d}}}}}$

Now let's solve it !

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✤ Solution :

↪ Refer to the second attachment...!

By applying the Newton 2nd law we can write that

• Net force = m.a

→ 2 × 10 - T = 2 × a [ While taking 10 the value of g ]

Now let's calculate tension of string

★ Tension = 20 - 2a ...(1)

We can see that in order to calculate tension we need to find the acceleration .so ,

★ Acceleration = v² = u² + 2as

Now let's write the value

• v ( Final velocity ) = 1

• u ( Initial velocity ) = 0 [ initial velocity was in rest therefore it will be zero ]

• s ( Displacement ) = 25 cm = 25/100 m

Let's put the value

★ Acceleration = v² = u² + 2as

➙ (1)² = (0)² + 2 × a × 25/100

➙ a = 2m/s²

Now , we know the value of acceleration . so , let's put the value of acceleration in order to find tension

★ Tension = 20 - 2a ...(1)

➙ 20 - 2 × 2

➙ 16 Newton

We know the formula of work done which is ,

$: \Longrightarrow{{\bf{{W=\overrightarrow{F}\cdot\overrightarrow{d}}}}}$

we know that force (tension) is in upward direction and displacement is in downward direction .

$: \Longrightarrow{{\sf{{W= |\overrightarrow{F}| \times |\overrightarrow{d}|. \cos180 \degree }}}}$

Now let's write the value

• F ( force (tension) ) = 16

• D ( Displacement ) = 1/4

• Cos 180° = -1

let's put the value,

$: \Longrightarrow{{\sf{{W= 16 \times \dfrac{1}{4} \times ( - 1 )}}}}$

$: \Longrightarrow{{\sf{{W= \blue{- 4 \: joule}}}}}$

☀️ Hence, solved !!

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