Question

# A force of $5N$ produces an acceleration of $8m{s}^{-2}$ on a mass ${m}_{1}$ and an acceleration of $24m{s}^{-2}$ on a mass ${m}_{2}$. What acceleration (in $m{s}^{-2}$) would the same force provide if both the masses are tied together?

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Solution

## Step1: Given dataA force of $5N$ produces an acceleration of $8m{s}^{-2}$ on a mass ${m}_{1}$ and acceleration of $24m{s}^{-2}$ on a mass ${m}_{2}$.Step2: Formula used$F=m×a\left[whereF=force,m=mass,a=acceleration\right]$Step3: Calculating accelerationForce is defined as the product of mass and acceleration. $F=m×a$Here, the force applied is the same on masses ${m}_{1}and{m}_{2}$Mass will be obtained by dividing the force with the acceleration produced by the mass. $m=\frac{F}{a}$Mass ${m}_{1}$ is obtained by dividing the force of $5N$ with the acceleration of $8m{s}^{-2}$${m}_{1}=\frac{F}{{a}_{1}}\phantom{\rule{0ex}{0ex}}{m}_{1}=\frac{5}{8}kg\left[{a}_{1}=8m{s}^{-2},F=5N\right]$Mass ${m}_{2}$ is obtained by dividing the force of $5N$ with the acceleration of $24m{s}^{-2}$${m}_{2}=\frac{F}{{a}_{2}}\phantom{\rule{0ex}{0ex}}{m}_{2}=\frac{5}{24}kg\left[{a}_{2}=24m{s}^{-2},F=5N\right]$The resultant mass when both the masses are tied together will be $M={m}_{1}+{m}_{2}\phantom{\rule{0ex}{0ex}}M=\frac{5}{8}+\frac{5}{24}\left[{m}_{1}=\frac{5}{8}kg,{m}_{2}=\frac{5}{24}kg\right]\phantom{\rule{0ex}{0ex}}M=\frac{20}{24}\phantom{\rule{0ex}{0ex}}M=\frac{5}{6}kg$The mass resulting when the two masses are combined is $\frac{5}{6}kg$.Acceleration (in $m{s}^{-2}$ ) when both the masses are tied together is$a=\frac{F}{m}\phantom{\rule{0ex}{0ex}}a=\frac{5}{\frac{5}{6}}\left[F=5N,m=\frac{5}{6}kg\right]\phantom{\rule{0ex}{0ex}}a=6m{s}^{-2}$Hence, the acceleration produced when the masses are tied together with the same force will be $6m{s}^{-2}$.

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