Question

#Quality answer

★Brainly star

★Brainly moderator, and other best user.

$\sf\small\green{Question:-}$
A mass Is lifted to a height in 10 sec and if the same mass is lifted to the same height in 20 sec the work done in the two cases in the ratio​.

⚠️☠️ website Maharaj se Uttar chaapne ki koshish Na Karen⚠️☠️​

Answer 1

Answer:

1:1 is the required answer .

Explanation:

According to the Question

It is given that,

• Time ,T1 = 10s
• Time ,T2 = 20s

Let the mass of the body be m kg.

The the height lifted by body be h metres .

we have to calculate the work done in these two cases.

Here, both body is lifted upward and reach to height h m.

So , we calculate the work done by the body in against gravity. Acceleration due to gravity is same in both cases .

• W = mgh

Acceleration due to gravity is 10m/s².

Work Done in 1st case

→ W = mgh

→ W = 10mh ---------(I)

Work Done in 2nd case :

→ W' = mgh

→ W' = 10mh--------(ii)

On comparing equation (i) & (ii) we get

→ W/W' = 10mh/10mh

→ W/W' = 1/1

• Hence, the ratio of work done in these two cases is 1:1.

From the above solution we observe that there is no dependency of time in work done .

Answer 2

Answer:

1:1

Solution:

lets assume all forces are conservative

work done can be stated as change in potential energy or ∆U

let's take the earth surface or initial position of object to be detum

So in first case

initial potential energy is 0

Final potential energy is mgh

hence work done in first case is ∆U=mgh

In second case

initial potential energy is 0

Final potential energy is ∆U=mgh

hence in second case work done is ∆U=mgh

Hence in both cases work done is same so the ratio will be

$\mathtt{ \frac{mgh}{mgh} } = \frac{1}{1}$

In the above solution one thing is noticed that work done in this case in independent of time which means if a particle travel any distance in any time the workdone will be same.

$\huge\boxed{ \mathtt{answer = 1:1}}$