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# a body is dropped from a height h when loss in its potential energy

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## A body is dropped from a height h . When loss in its potential energy is U then its velocity is v . The mass of the body is:

Click here👆to get an answer to your question ✍️ A body is dropped from a height h . When loss in its potential energy is U then its velocity is v . The mass of the body is:

Question

## A body is dropped from a height h. When loss in its potential energy is U then its velocity is v. The mass of the body is:

A

2v U 2 ​

B

U 2v ​

C

U 2 2v ​

D

v 2 2U ​ Medium Open in App

Updated on : 2022-09-05

Solution Verified by Toppr

Correct option is D)

loss in PE=gain in KE

U= 2 1 ​ mv 2 m= v 2 2U ​

53 9

स्रोत : www.toppr.com

## " A body is dropped form a height h.When loss in its potential energy is "'U'," its velocity is "V" .The mass of the body is "

2U/V^2" A body is dropped form a height h.When loss in its potential energy is "'U'," its velocity is "V" .The mass of the body is "

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" A body is dropped form a hei...

" A body is dropped form a height h.When loss in its potential energy is "'U'," its velocity is "V" .The mass of the body is "

Updated On: 27-06-2022

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Text Solution Open Answer in App Solution 2U/V^2

Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.

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## A body is dropped from a height \$h\$. When the loss in its potential energy is \$U\$ then its velocity is \$v\$. The mass of the body is:A) \$\\dfrac{{{U^2}}}{{2v}}\$B) \$\\dfrac{{2v}}{U}\$C) \$\\dfrac{{2v}}{{{U^2}}}\$D) \$\\dfrac{{2U}}{{{v^2}}}\$

A body is dropped from a height \$h\$. When the loss in its potential energy is \$U\$ then its velocity is \$v\$. The mass of the body is:A) \$\\dfrac{{{U^2}}}{{2v}}\$B) \$\\dfrac{{2v}}{U}\$C) \$\\dfrac{{2v}}{{{U^2}}}\$D) \$\\dfrac{{2U}}{{{v^2}}}\$. Ans: Hint The ...

A body is dropped from a height

h h

. When the loss in its potential energy is

U U

then its velocity is

v v

. The mass of the body is:

A) U 2 2v U22v B) 2v U 2vU C) 2v U 2 2vU2 D) 2U v 2 2Uv2 Answer Verified 222k+ views

Hint The law of conservation of energy relates the body’s kinetic and potential energy. When the body is at a height

h h

, it will only have potential energy and when it's falling and has lost potential energy

U U

it will have some kinetic energy. We will use this relation to equate the loss in potential energy of the body with the gain in kinetic energy of the body as it accelerates downwards.

In this solution, we will use the following formula

Kinetic energy of a body:

K= 1 2 m v 2 K=12mv2 where m m is its mass and v v is its velocity.

The law of conservation of energy tells us that energy cannot be created or destroyed but it can only be converted from one form to the other.

We’ve been given that a body is dropped from a height

h h

. When the loss in its potential energy is

U U

, the amount of energy will have converted into kinetic energy. We know that the kinetic energy of a body travelling with velocity

v v and mass m m

can be calculated as:

⇒K= 1 2 m v 2 ⇒K=12mv2

So, the loss in the potential energy

U U

is converted into kinetic energy so we can write

⇒U= 1 2 m v 2 ⇒U=12mv2

Dividing both sides by

v 2 /2 v2/2 , we get m= 2U v 2 m=2Uv2

which corresponds to option (D).

Note

Here we have assumed that there is no air resistance however in an actual scenario, air resistance will decrease the kinetic energy of the body due to frictional losses. We can also select the correct option since only option (D) has the dimensions of mass while the rest of the options have different dimensions which are not the dimensions of mass.

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Mohammed 7 day ago

Guys, does anyone know the answer?