Solveeit Logo

Question

Question: How do you solve for m in the equation \(\dfrac{1}{2}m{{v}^{2}}=\dfrac{1}{2}k{{x}^{2}}\) ?...

How do you solve for m in the equation 12mv2=12kx2\dfrac{1}{2}m{{v}^{2}}=\dfrac{1}{2}k{{x}^{2}} ?

Explanation

Solution

To solve for m in the equation 12mv2=12kx2\dfrac{1}{2}m{{v}^{2}}=\dfrac{1}{2}k{{x}^{2}} , we have to cancel the common terms from both the sides. Then, we have to take v2{{v}^{2}} to the RHS. We can find the value of m clearly, if v=xv=x .

Complete step-by-step solution:
We have to solve for m in the equation 12mv2=12kx2\dfrac{1}{2}m{{v}^{2}}=\dfrac{1}{2}k{{x}^{2}} . Let us cancel 12\dfrac{1}{2} from both the sides.
\requirecancel12mv2=\requirecancel12kx2\Rightarrow \require{cancel}\cancel{\dfrac{1}{2}}m{{v}^{2}}=\require{cancel}\cancel{\dfrac{1}{2}}k{{x}^{2}}
We will get the result of the above simplification as
mv2=kx2...(i)\Rightarrow m{{v}^{2}}=k{{x}^{2}}...\left( i \right)
Now, we have to take v2{{v}^{2}} to the RHS. This will be the divisor in the RHS.
m=kx2v2\Rightarrow m=\dfrac{k{{x}^{2}}}{{{v}^{2}}}
The value of m depends on the value of x. Let us consider equation (i).
If v2=x2{{v}^{2}}={{x}^{2}} or v=xv=x , we can find the value of m as
mv2=kv2\Rightarrow m{{v}^{2}}=k{{v}^{2}}
Let us cancel v2{{v}^{2}} from both the sides.
m\requirecancelv2=k\requirecancelv2\Rightarrow m\require{cancel}\cancel{{{v}^{2}}}=k\require{cancel}\cancel{{{v}^{2}}}
We will get the result of the above simplification as
m=k\Rightarrow m=k
Therefore, the value of m is kx2v2\dfrac{k{{x}^{2}}}{{{v}^{2}}} and if v=xv=x , we will get m=km=k .

Note: Students must know to solve algebraic equations and the rules associated with it. When we move a positive term to other side, it becomes negative. Likewise, when we move a negative term to other side, it will become positive. When we move a divisor to one side, it will be multiplied with the terms on the moved side. We cannot find the value of m unless the values of k,x and v are given.The given equation is related to physics, that is, mass and energy of a spring. The kinetic energy of the spring is equal to its elastic potential energy.