Question

The kinetic energy of a body is increased most by doubling its:
(A) mass
(B) weight
(C) speed
(D) density

Answer 1

Hint: The kinetic energy of a body is given by the product of its mass and velocity squared upon two. We shall mathematically break it down in the form of parameters mentioned in the options. These parameters are, mass, weight, speed and density. After that we will find the new kinetic energy in each case and compare them to see, in which case we get the maximum kinetic energy.

Complete answer:
We will first assign these parameters some known terms so that it's easier to use in our calculations.
Let the mass of the body be denoted by ‘M’.
Then the weight of the body (say W) will be equal to:
$\Rightarrow W=Mg$
Now, let the speed of the body be given by the term ‘V’.
And the density of the body is given by the term $\rho $ .
The initial kinetic energy of the body is equal to:
$\Rightarrow K.E{{.}_{I}}=\dfrac{1}{2}M{{V}^{2}}$
Now, in the first case the mass has to be doubled. Therefore, the new mass will be 2M.
Thus, the new kinetic energy will be equal to:
$\Rightarrow K.E{{.}_{1}}=\dfrac{1}{2}(2M){{V}^{2}}$
$\Rightarrow K.E{{.}_{1}}=M{{V}^{2}}$ [Let this expression be equation number (1)]
Now, in the second case, the weight is doubled. Therefore, the new weight is 2Mg.
This condition is the same as doubling the mass. Therefore, the kinetic energy in this case will be equal to:
$\Rightarrow K.E{{.}_{2}}=M{{V}^{2}}$ [Let this expression be equation number (2)]
Now, in the third case, the speed of the body is doubled. Thus, the new speed will be equal to 2V. Thus, the new kinetic energy will be equal to:
$\Rightarrow K.E{{.}_{3}}=\dfrac{1}{2}M{{(2V)}^{2}}$
$\Rightarrow K.E{{.}_{3}}=2M{{V}^{2}}$ [Let this expression be equation number (3)]
Now, in the last case the density of the body under observation is doubled, keeping the volume constant. So, the mass of the object has also doubled. Thus, this case is also the same as first case. Hence, the kinetic energy in this case will be equal to:
$\Rightarrow K.E{{.}_{4}}=M{{V}^{2}}$ [Let this expression be equation number (4)]
On comparing the final kinetic energy of the body in all the four cases, we see that the kinetic energy of the body is increased most when its speed is doubled.

Hence, option (C) is the correct option.

Note:
In all the four cases, the kinetic energy of the body increased. This is because the parameters that were increased were directly proportional to the formula of kinetic energy. If these were decreased by the same factor, then kinetic energy would be the least upon decreasing the speed.