Question

Please answer fast ​

Answer 1

Answer:

Given:

Block of mass "m" is pulled by a string at a constant acceleration of (g/3) m/s².

To find:

Work done by tension

Diagram:

First draw the Free-body diagram (FBD) of the block. Refer to the attached photo to understand better.

Calculation :

As per the FBD,

T - mg = ma

=> T - mg = m(g/3)

=> T = mg + m(g/3)

=> T = (4/3)mg

Now , work done = force × displacement

=> W = {(4/3)mg} × h

=> W = (4/3)mgh.

So the answer is option 4)

Answer 2

Question :-

A string is used to pull a block of mass m vertically up by a distance h at a constant acceleration g/3 .Then find the work done by tension in the string .

Answer :-

$\red{\boxed{\green{ \boxed{ \pink{\sf{work \: = \frac{4mgh}{3} }}}}}} \:$

To Find :-

→ Work done by tension in the string .

Formula used :-

$\sf{second \: law \: of \: newton \: } \\ \rightarrow \boxed{\sf{ F\: = \: ma \: }}$

Here two force applied ,

• Tension
• Force by block

$\rightarrow \boxed{ \pink{\sf{work \: = force \times displacement \:} }}$

Step - by - step explanation :-

Given that ,

• Mass of block is m .
• acceleration is a .
• Displacement is h .

Solution :-

Applied second law of Newton ,

$\rightarrow \sf{F = m \: a} \\ \\ \star \: \sf{ force \: by \: string \: = T }\\ \\ \star \: \sf{force \: by \: block \: = \: mg} \\ \\ \rightarrow \sf{ T \: - mg \: = \: m \: a \: } \\ \\ \because \: \sf{a \: = \frac{g}{3} } \\ \\ \therefore \: \\ \rightarrow \sf{T \: - mg \: = \: m \: \frac{g}{3} } \\ \\ \rightarrow \sf{ T \: = \frac{mg}{3} + mg} \\ \\ \rightarrow \boxed{ \red{\sf{T \: = \frac{4mg}{3}} }} \\ \\ \sf{force \: by \: string \: (T) = \red{ \frac{4mg}{3} }}$

_____________________________

We know that ,

$\star \: \pink{\sf{work \: } = \green{ force \times displacement \:} } \: \\ \\ \rightarrow \sf{ work \: = \frac{4mg}{3} \times h} \\ \\ \rightarrow \boxed{\green{ \boxed{ \pink{\sf{work \: = \frac{4mgh}{3} }}}}}$

Option (4) is correct .

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