
Consider a drop of rainwater having mass $1g$ falling from a height of $1km$. It hits the ground with a speed of $50m/s$. Take '$g$' constant with a value $10m/{s^2}$. The work is done by the $\left( i \right)$ gravitational force $\left( {ii} \right)$ the resistive force of airs is:
$\left( a \right)$ $\left( i \right)$ $10J$ $\left( {ii} \right)$ $ - 8.75J$
$\left( b \right)$ $\left( i \right)$ $10J$ $\left( {ii} \right)$ $8.75J$
$\left( c \right)$ $\left( i \right)$ $1.25J$ $\left( {ii} \right)$ $ - 8.75J$
$\left( d \right)$ $\left( i \right)$ $100J$ $\left( {ii} \right)$ $ - 8.75J$
Answer
126k+ views
Hint First of all for this question we will calculate the work done by the gravitational force by using the formula $Wg = mgh$ and then we will find out the change in the kinetic energy and when we get these two values, by applying the work-energy theorem we will get the resistive force of the air.
Formula used:
Work done by the gravitational force will be given by
$Wg = mgh$
Here,
$Wg$, will be the work done duo gravitational force
$m$, will be the mass
$g$, will be the acceleration due to gravity
$h$, will be the height
Complete Step By Step Solution
Firstly we will see the values which are given to us.
$m = 1g = \dfrac{1}{{1000kg}}$
$h = 1km = 1000m$
Presently by applying the equation
Work done by the gravitational power will be given by
$Wg = mgh$
Substituting the values, we get
$ \Rightarrow Wg = \dfrac{1}{{1000}} \times 10 \times 1000$
On understanding the above condition, we get
$ \Rightarrow Wg = 10J$
Therefore, the$10J$ work is done by the gravitational force.
Now we will see the change in the kinetic energy
And it will be given by
$\vartriangle K.E = \dfrac{1}{2}m{v^2} - 0$
Now we will substitute the values, we get
$ \Rightarrow \dfrac{1}{2}\dfrac{1}{{1000}} \times 50 \times 50$
Therefore on solving the above equation, we get
$ \Rightarrow 1.25J$
Now by using the work-energy theorem,
We know
Work energy theorem$ = {W_g} + {W_{air}} = \vartriangle K.E$
So now on substituting the values, we get
$ \Rightarrow 10J + res{\text{istance}} = 1.25 - 10$
From here we will calculate the value of resistance,
And it will be equal to
Resistance$ = - 8.75J$
Hence the option $A$ is correct.
Note The work-energy hypothesis expresses that the work done by all powers following up on a molecule approaches the adjustment in the molecule's active energy. The work-energy guideline expresses that the difference in energy of a framework is equivalent to work done on or by the framework. I don't generally know a thorough confirmation however this standard depends on the law of preservation of energy, which is one the key and significant laws that are utilized wherever in material science. Energy when any cooperation, impact, and so on are consistently the same.
Formula used:
Work done by the gravitational force will be given by
$Wg = mgh$
Here,
$Wg$, will be the work done duo gravitational force
$m$, will be the mass
$g$, will be the acceleration due to gravity
$h$, will be the height
Complete Step By Step Solution
Firstly we will see the values which are given to us.
$m = 1g = \dfrac{1}{{1000kg}}$
$h = 1km = 1000m$
Presently by applying the equation
Work done by the gravitational power will be given by
$Wg = mgh$
Substituting the values, we get
$ \Rightarrow Wg = \dfrac{1}{{1000}} \times 10 \times 1000$
On understanding the above condition, we get
$ \Rightarrow Wg = 10J$
Therefore, the$10J$ work is done by the gravitational force.
Now we will see the change in the kinetic energy
And it will be given by
$\vartriangle K.E = \dfrac{1}{2}m{v^2} - 0$
Now we will substitute the values, we get
$ \Rightarrow \dfrac{1}{2}\dfrac{1}{{1000}} \times 50 \times 50$
Therefore on solving the above equation, we get
$ \Rightarrow 1.25J$
Now by using the work-energy theorem,
We know
Work energy theorem$ = {W_g} + {W_{air}} = \vartriangle K.E$
So now on substituting the values, we get
$ \Rightarrow 10J + res{\text{istance}} = 1.25 - 10$
From here we will calculate the value of resistance,
And it will be equal to
Resistance$ = - 8.75J$
Hence the option $A$ is correct.
Note The work-energy hypothesis expresses that the work done by all powers following up on a molecule approaches the adjustment in the molecule's active energy. The work-energy guideline expresses that the difference in energy of a framework is equivalent to work done on or by the framework. I don't generally know a thorough confirmation however this standard depends on the law of preservation of energy, which is one the key and significant laws that are utilized wherever in material science. Energy when any cooperation, impact, and so on are consistently the same.
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