
What does $9.8{\text{ }}N/kg$ mean?
Answer
438.3k+ views
Hint: We can use dimensional analysis to compare the given parameter with the parameters available to us. Also, we can observe the units carefully and use a known formula to describe this parameter and then answer this question.
Formula Used:
$F{\text{ }} = {\text{ }}ma$
Where, $F$ is the force on a body, $m$ is the mass of the body and $a$ is the acceleration of the body.
${F_G} = \dfrac{{GMm}}{{{R^2}}}$
Where, $G$ is the universal gravitational constant, $M$ is the mass of the planet, $m$ is the mass of the object and $R$ is the distance between the planet and object.
${F_G} = mg$
Where, $g$ is the acceleration due to gravity.
Complete step by step solution:
The unit of the given parameter is $N/kg$.
We know the dimension of $N$ is $\left[ {ML{T^{ - 2}}} \right]$ and the dimension of $kg$ is $\left[ M \right]$. Thus, the net dimension of the parameter is $\dfrac{{\left[ {ML{T^{ - 2}}} \right]}}{{\left[ M \right]}}{\text{ }} = {\text{ }}\left[ {L{T^{ - 2}}} \right]$.
We know this is the dimension of acceleration.
Now,
Again,
The units involved in the parameter are $N$ and $kg$ which are the units of force and mass respectively out of which mass is an independent quantity.
Thus, we will use the formula of force
$F{\text{ }} = {\text{ }}ma$
Further, we can write
$a{\text{ }} = {\text{ }}\dfrac{F}{m}$
Thus,
The given parameter is acceleration.
Hence, $9.8{\text{ }}N/kg$ is just $9.8{\text{ }}m/{s^2}$ which is acceleration and speaking more precisely is the acceleration due to gravity of earth.
Additional Information:
We know that the force on an object by a planet is given by
${F_G} = \dfrac{{GMm}}{{{R^2}}}$
Also,
Force of the gravity on an object is given by
${F_G} = mg$
Now,
We equate the two
$\dfrac{{GMm}}{{{R^2}}} = mg$
Further, we get
$\dfrac{{GM}}{{{R^2}}} = g$
Thus,
The equation for the acceleration due to gravity of the planet is
$g = \dfrac{{GM}}{{{R^2}}}$
Note: Students should be careful while performing the dimensional analysis as the concept is a simple one but under the circumstances of overconfidence, the students overlook the evaluation process. Students also make mistakes while substituting the dimensions of a parameter.
Formula Used:
$F{\text{ }} = {\text{ }}ma$
Where, $F$ is the force on a body, $m$ is the mass of the body and $a$ is the acceleration of the body.
${F_G} = \dfrac{{GMm}}{{{R^2}}}$
Where, $G$ is the universal gravitational constant, $M$ is the mass of the planet, $m$ is the mass of the object and $R$ is the distance between the planet and object.
${F_G} = mg$
Where, $g$ is the acceleration due to gravity.
Complete step by step solution:
The unit of the given parameter is $N/kg$.
We know the dimension of $N$ is $\left[ {ML{T^{ - 2}}} \right]$ and the dimension of $kg$ is $\left[ M \right]$. Thus, the net dimension of the parameter is $\dfrac{{\left[ {ML{T^{ - 2}}} \right]}}{{\left[ M \right]}}{\text{ }} = {\text{ }}\left[ {L{T^{ - 2}}} \right]$.
We know this is the dimension of acceleration.
Now,
Again,
The units involved in the parameter are $N$ and $kg$ which are the units of force and mass respectively out of which mass is an independent quantity.
Thus, we will use the formula of force
$F{\text{ }} = {\text{ }}ma$
Further, we can write
$a{\text{ }} = {\text{ }}\dfrac{F}{m}$
Thus,
The given parameter is acceleration.
Hence, $9.8{\text{ }}N/kg$ is just $9.8{\text{ }}m/{s^2}$ which is acceleration and speaking more precisely is the acceleration due to gravity of earth.
Additional Information:
We know that the force on an object by a planet is given by
${F_G} = \dfrac{{GMm}}{{{R^2}}}$
Also,
Force of the gravity on an object is given by
${F_G} = mg$
Now,
We equate the two
$\dfrac{{GMm}}{{{R^2}}} = mg$
Further, we get
$\dfrac{{GM}}{{{R^2}}} = g$
Thus,
The equation for the acceleration due to gravity of the planet is
$g = \dfrac{{GM}}{{{R^2}}}$
Note: Students should be careful while performing the dimensional analysis as the concept is a simple one but under the circumstances of overconfidence, the students overlook the evaluation process. Students also make mistakes while substituting the dimensions of a parameter.
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