Question

In an experiment, mass of an object is measured by applying a known force on it, and then measuring its acceleration. If, in the experiment, the measured values of applied force and the measured acceleration are F=10.0$ \pm 0.2N$ and a=1.00$ \pm 0.01\dfrac{m}{{{s^2}}}$ , respectively, the mass of object is
A) 10.0Kg
B) 10.0$ \pm $ 0.1Kg
C) 10.0$ \pm $ 0.3Kg
D) 10.0$ \pm $ 0.4Kg

Answer 1

Hint:Here in the question we are given force as F=10.0$ \pm 0.2N$ and value of acceleration a=1.00$ \pm 0.01\dfrac{m}{{{s^2}}}$with errors. We all know the formula for force which is F=ma and for solving errors we only need to do is error in force divided by F is equal to error in m(mass) divided by mass added with error in acceleration divided by acceleration.

Step by step solution:
Step 1:
F=ma states that force is equal to the product of mass and acceleration of an object and clearly states in Newton's second law of motion 
Newton's second law of motion can be formally stated as follows: The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.
Step 2:
Now coming to the question, let’s see what are the things which we are already given
We are given force, F=10.0$ \pm 0.2N$ and
Acceleration a=1.00$ \pm 0.01\dfrac{m}{{{s^2}}}$ , respectively.
The mass of object is F=ma which is 10.0=m(1.0)
This gives a value mass is equal to 10kg
Now to calculate the error we need to apply log on both sides.
($\therefore \log a$ =$\dfrac{1}{a}$ ), this is the property of log function.
Applying log on both sides in F=ma, log F=log m + log a
This is equal to $\dfrac{{\Delta F}}{F}$ =$\dfrac{{\Delta m}}{m}$ +$\dfrac{{\Delta a}}{a}$
$\Delta F$is given =0.2, $\Delta m$ need to find out and $\Delta a$=0.01
$\dfrac{{\Delta m}}{m}$= $\dfrac{{\Delta F}}{F}$−$\dfrac{{\Delta a}}{a}$(for finding error in mass)
Putting value of mass (10kg) and values of force and acceleration with their respective error we will get r $\Delta m$ equals to 0.3
So the value of mass is equal to 10$ \pm 0.3Kg$

Hence option C is correct

Note:The standard error of a mean provides a statement of probability about the difference between the mean of the population and the mean of the sample. If we take the mean plus or minus three times its standard error. The smaller the standard error, the less the spread and the more likely it is that any sample mean is close to the population mean. A small standard error is thus a good thing.