Question

a stone of mass m is held at rest in water.The stone is released and falls vertically a distance h.the stone reaches a speed v.Which expression gives the work done against resistive force ?

Answer 1

Answer:

Given: Mass of stone = m

distance covered = h

velocity reached = v

To find : work done against Frictional force

Concept: In this case, the net Frictional force is provided by 2 forces namely

a) Buoyant force

b) Stroke's force

Calculation: Let the acceleration offered by Frictional force be "a"

Now applying equation of kinematics:

v² = u² + 2as

=> v² = 0² + 2×a×h

=> v² = 2ah

=> a = (v²/2h)

Now force experienced

= mass × acceleration

= m×a

= m (v²/2h)

Now work done = force ×displacement

W = m(v²/2h)× h

W = (mv²/2)

So the answer is [mv²/2].

Answer 2

Question :-

a stone of mass m is held at rest in water.The stone is released and falls vertically a distance h.the stone reaches a speed v.Which expression gives the work done against resistive force ?

Answer:-

$\boxed{\sf{work \: = \frac{1}{2}m \: {v}^{2} }} \:$

To Find :-

→ Find work done against Force .

Formula used :-

Here we used,

$\sf{ third \: equation \: of \: motion} \\ \\ \rightarrow \: \boxed{ \sf{ {v}^{2} = {u}^{2} + 2as}} \\ \\ \rightarrow \small \sf{ \boxed{\sf{work \: = force \: \times \: displacement}}} \\ \\ \rightarrow \small \sf{ \boxed{\sf{ force \: = \: mass \: \times \: acceleration}}}$

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Step - by - step explanation :-

Let, acceleration is "a",

Initial velocity ( u ) = 0,

Displacement is "h",

Now applying the third equation of motion ,

$\rightarrow \: \sf{ {v}^{2} = 0 + 2a \times h }\: \\ \\ \rightarrow \: \sf{a \: = \frac{ {v}^{2} }{2h} } \: \: \: ......(1)$

Now find force to finding the work done ,

$\rightarrow \: \sf{force \: = m \: \times a }\\ \\ \rightarrow \: \sf{ force \: = m \: \times \frac{ {v}^{2} }{2h} } \: \: \: ....(2)$

Now required work done is ,

$\boxed{ \sf{work \: = force \: \times displacement \: }} \\ \\ \bf{ from \: (2)} \\ \\ \rightarrow \: \sf{work \: = m \: \frac{ {v}^{2} }{2h} \times h} \\ \\ \rightarrow \: \boxed{\sf{work \: = \frac{1}{2}m \: {v}^{2} }}$

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