Experiment 7

Newton's Second Law of Motion


The objective is to experimentally verify Newton's Second Law.


A computer with Internet connection, a calculator, paper, and pencil

Theory:  Newton's 2nd law states that

a nonzero net force ΣF acting on mass M accelerates it such that ΣF = Ma    or     ΣF/M = a.  

In the course of experiment, you will show that if the ratio of the net force, ΣF, to the total mass Mtotal  is kept constant, acceleration must also remain constant.

To verify that  ΣF/M = a  is valid, calculate it from:  x = (1/2)at2 + vi t as well to show that both a's will result in the same value.      

In Fig. 1, assuming no friction, the hanging mass m causes both masses to move and accelerate.  The force of gravity on m is mg.   This force moves a total mass of m + M.  With no friction, mg is the only force causing motion.   In such case mg is the net force or ΣF.


Using Newton's 2nd Law:

 ΣF = (Mass)(Accel.) , we may write:

mg  = (m + M)a   or, 

a = mg /(m + M).


Figure 1

In Fig. 2, with friction present, the opposing force of friction must be included in ΣF.  Since N = w in magnitude, we may write or N = Mg.  

The force of kinetic friction is  Fk = μN , or  Fk = μMg.   Here the net force becomes:  ΣF = mg - μMg   and the total mass to be moved is still m+M. 

Using Newton's 2nd Law:

 ΣF = (Mass)(Accel.)

mg - μMg = (m + M)a   or, 

a = (mg - μMg )/(m + M).


Figure 2:



If you have not done so, you may need to add the following Website to your Java exception list: http://www.walter-fendt.de/

To do this, follow the path (Windows operating system),

Start → All Programs → Java → Configure Java → Security (use High) → Edit Site List … → Add → Type in the site URL (http://www.walter-fendt.de/).

The method is to solve for a from ΣF = Ma and call it the measured value and also find the same a from s = (1/2)at2 +vit  and call it the accepted value and see how close they turn out.   With vi = 0 in all cases,  s = (1/2)at2 and solving for a we get: a = 2s/t2.  The applet uses this formula each time to calculate the accepted value.

Click on the following link: http://www.walter-fendt.de/ph14e/n2law.htm .   Note that masses in the applet are in (grams).  For ease of calculation, suppose they are given in kilograms (kg) and ignore the grams.   This is not going to affect the results because as you see in both formulas the units for masses cancel in the numerator and the denominator of each equation.

Part A: No Friction ( μ = 0 )

Referring to Table 1, for the first 9 trials, the coefficient of friction μ must be set equal to zero.  After each run, the applet must be reset.  

1) Run the applet for the values given in the first row.  In the first row, both M and m are given.   Use the formula a = mg /(m + M) under Fig. 1 and calculate the "Measured a"  and record it in Table 1.

2) Record the "Accepted a" that the applet calculates from a = 2s/t2  in Table 1 in the space provided.   Calculate the % error and record it in Table 1 in the last column.

3) For rows 2 through 9, in each row, use the given M and guess values for m until you get the same acceleration as you obtained for the first row.  The applet calculates the accepted value of acceleration for you, each time.  The goal is to arrive at the same acceleration for all 9 cases of Part A.  This can be achieved by changing m in each row until the accepted a and measured a are close enough to result in a percent error under 2 or 3 percent and at the same time very close to the values of a of the previous row (s).

Part B:  Friction Present

Include the given coefficients of kinetic friction in each case.  Follow the same procedure as applied to Part AAgain, whatever value of a you obtain for the first row, you should change m in all other 8 cases of Part B such that you get the same acceleration for all of them as you got in the first row.  A percent error under 2 percent is acceptable.


Given:       g = 9.81 m/s2.


Table 1

Trial M m μk ΣF =

mg - μMg

Total mass

M + m

Measured a =


Accepted a =

 2s /t2



Part A (kg) (kg)   ( N ) (kg) ( m/s2 ) ( m/s2 )  
1 100.0 4.00 0          
2 150.0   0          
3 200.0   0          
4 250.0   0          
5 300.0   0          
6 350.0   0          
6 400.0   0          
8 450.0   0          
9 500.0   0          
Part B                
1 100.0 11.73 0.080
2 150.0 0.080
3 200.0 0.080
4 250.0 0.100
5 300.0 0.100
6 350.0 0.100
7 400.0 0.160
8 450.0 0.160
9 500.0 0.160




Show a sample calculation.

Comparison of the results:

Provide the percent error formula used.


State your conclusions of the experiment with reference to the values obtained in columns 5, 6, 7, 8, and 9 of the above Table.


Provide a discussion, if necessary.