Question

For each trial compare the slope of the regression line to the mass being accelerated. What does the slope represent? What is the percentage error between the two values?

Trial one

mass: .0659 Kg

Regression Line for force vs. acceleration: F= 0.5771 N/m/s^2 - 0.02760 N

Trial two

Mass: .5659

Regression Line for force vs acceleration: F= 1.100 N/m/s^2 - 0.07264

Answer 1

As per newton's second law

F = ma

m= F/a

Thereofore,

slope of force vs acceleration represents mass

In trial one, the slope you are getting is 0.5771 which is the mass

Therefore,

percentage error = (value from the graph - theoretical value / theoetical value) * 100

The mass which you gave for trial mass seems to be so small that percentage error would be extremely high when compared to value obtained through graph. Please re-check the value

Now, for second trial,

You had theoretical mass as 0.5659 Kg and the mass which you got from graph is 1.1

Therefore,

percentage error = (|1.1 - 0.5659| / 0.5659 )* 100 = 94 % (such a big error). This cannot be possible.

Anyway, the method to find the percentage error will be same. Either you values of mass are wrong. Please note that here you need to consider mass of the system as theoretical mass.