Work done in raising a body of mass 2 kg from ground level to a height 5 m is given by $\left( {g = 10m/{s^2}} \right)$:
A) 10 J
B) 100 J
C) 0 J
D) 2.5 J
Answer
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Hint: As the body moves from one position to another, it will suffer the change in its potential energy. This change in potential energy is equal to the work done and thus we can calculate its magnitude from here. Potential energy is given by the product of mass, height and acceleration due to gravity.
Complete step by step answer:
When the object was at ground level, its height was zero and then it was lifted to another level at height of 5 m (given). The body will suffer a change in energy due to its position, so, there will be change in its potential energy.
Now, this change in potential energy is equal to the work done by the object in moving from ground to higher level.
$WD = P{E_2} - P{E_1}$ ___________ (1)
Potential energy is given as:
$PE = mgh$
For both the cases it can be given as:
When the object is at ground level:
$P{E_1} = mgh$, here
The height of the object will be 0 as it is placed on the ground, so when 0 is multiplied to other numbers the result will also be 0.
$ \Rightarrow P{E_1} = 0$
When the object is lifted:
$P{E_2} = mgh$, here
Mass of the body (m) = 2 Kg (given)
Acceleration due to gravity (g) = $10m/{s^2}$
Height from the ground level (h) = 5 m (given)
\[
\Rightarrow P{E_2} = 2 \times 10 \times 5 \\
\Rightarrow P{E_2} = 100 \\
\]
Substituting the values in (1), we get:
$
WD = 100 - 0 \\
\Rightarrow WD = 100J \\
$
As the SI Unit of work is joules (J).
Therefore, the work done in raising a body of mass 2 kg from ground level to a height 5m is given by 100 J
So, the correct answer is “Option B”.
Note:
We are always required to give the required answer in SI Units.
We say that change in potential energy is equal to the work done according to the conservation of energy. The energy does not get lost but gets converted into the energy.
To find the change in potential energy, we always subtract the energy at the initial position from the final position
Complete step by step answer:
When the object was at ground level, its height was zero and then it was lifted to another level at height of 5 m (given). The body will suffer a change in energy due to its position, so, there will be change in its potential energy.
Now, this change in potential energy is equal to the work done by the object in moving from ground to higher level.
$WD = P{E_2} - P{E_1}$ ___________ (1)
Potential energy is given as:
$PE = mgh$
For both the cases it can be given as:
When the object is at ground level:
$P{E_1} = mgh$, here
The height of the object will be 0 as it is placed on the ground, so when 0 is multiplied to other numbers the result will also be 0.
$ \Rightarrow P{E_1} = 0$
When the object is lifted:
$P{E_2} = mgh$, here
Mass of the body (m) = 2 Kg (given)
Acceleration due to gravity (g) = $10m/{s^2}$
Height from the ground level (h) = 5 m (given)
\[
\Rightarrow P{E_2} = 2 \times 10 \times 5 \\
\Rightarrow P{E_2} = 100 \\
\]
Substituting the values in (1), we get:
$
WD = 100 - 0 \\
\Rightarrow WD = 100J \\
$
As the SI Unit of work is joules (J).
Therefore, the work done in raising a body of mass 2 kg from ground level to a height 5m is given by 100 J
So, the correct answer is “Option B”.
Note:
We are always required to give the required answer in SI Units.
We say that change in potential energy is equal to the work done according to the conservation of energy. The energy does not get lost but gets converted into the energy.
To find the change in potential energy, we always subtract the energy at the initial position from the final position
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