You visited us 1 times! Enjoying our articles? Unlock Full Access!

Relation between Work Done and PE

A particle of...

Question

A particle of mass $5 kg$ is dropped from a tower of height $10 m$ . Find the difference of work done by gravity from point $A$ to $B$ $(W_{A B})$ and change in potential energy from $A$ to $B$ $(U_{A B})$ .

10 g h

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

5 g h

No worries! We‘ve got your back. Try BYJU‘S free classes today!

- 10 g h

No worries! We‘ve got your back. Try BYJU‘S free classes today!

- 5 g h

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $10 g h$
Work done by gravity from $A$ to $B$ :
$W_{A B} = m g h$
(+ve because displacement is in the direction of force)

Change in potential energy from $A$ to $B$
$U_{A B} = U_{B} - U_{A}$
$= m g (10 - h) - m g (10)$
$= - m g h$

$∴ W_{A B} - U_{A B}$
$= m g h - (- m g h)$
$= 2 m g h$
$= 2 \times 5 \times g h$
$= 10 g h$

Suggest Corrections

Similar questions

Q. A body of mass

5 kg

is dropped from a tower of height

10 m

. Find the difference between work done

(J)

by gravity and change in potential energy of the body from

^{'} A^{'}

^{'} B^{'}

(Take

g = 10 \frac{m}{s^{2}}

)

Q. A particle of mass

10 kg

is dropped from tower of height

10 m

. Find the ratio of workdone by gravity from point

^{'} A^{'}

^{'} B^{'}

and change in potential energy from

^{'} A^{'}

^{'} B^{'}

Q. A particle of mass

10 kg

is dropped from tower of height

10 m

. Find the ratio of workdone by gravity from point

^{'} A^{'}

^{'} B^{'}

and change in potential energy from

^{'} A^{'}

^{'} B^{'}

Q. A body of mass

5 k g

is taken from a height of

5 m

10 m

. Find the increase in its potential energy. Take

g = 10 m s^{- 2}

A ball of mass 2 kg is raised to the height of 60 m and then dropped from the top. In this case, the quantity which has different magnitude from the other given quantities is

Join BYJU'S Learning Program

A particle of mass 5 kg is dropped from a tower of height 10 m. Find the difference of work done by gravity from point A to B (WAB) and change in potential energy from A to B (UAB).

The correct option is A 10ghWork done by gravity from A to B: WAB=mgh (+ve because displacement is in the direction of force) Change in potential energy from A to B UAB=UB−UA =mg(10−h)−mg(10) =−mgh ∴WAB−UAB =mgh−(−mgh) =2mgh =2×5×gh =10gh

A particle of mass $5 kg$ is dropped from a tower of height $10 m$ . Find the difference of work done by gravity from point $A$ to $B$ $(W_{A B})$ and change in potential energy from $A$ to $B$ $(U_{A B})$ .