Question

Calculations and Data:
Angle: As measured by protractor:7.63 degrees
As measured by using inverse sin:7.4 degrees
Percent error:
((True value-experimental value)/true value))*100=
((7.4-7.63)/7.4)*100=-3.1081%
Time:
1.53s
1.4s
1.53s
1.44s
1.47s
Average time= (total of times when added/5)=1.492s
Sides: measured directly
A: mgh = (8.7kg) (9.81m/s2)(.1492m)(sin7.5)= 1.662*J
Vf=
square root(2*g*h)
square root(2*9.81m/s2*.1492m*sin7.5)
0.6161m/s

We did a lab in class where we rolled a ball (mass of 8.7kg) down an incliend plane. My group had our incline at 7.4 degrees above horizontal, with a hypotenuse of 115.57cm, opposite side of 14.92cm, and a length of 114.602cm. We calculated (perhaps incorrecly) that our acceleration was mgh = (8.7kg) (9.81m/s2)(.1492m)(sin7.5)= 1.662*J. Note we converted the cm units to meters. I was trying to find a way to change the angle (and thus the distance across from the angle (the hieght)) in order to show that the acceleration of the object gets larger with a greater angle. I can't find a way of doing this.

Answer 1

Comments:
I agree with percent error.
On average , how can you get more significant digits that you started with?

Your measurments of length, to the thousanth of cm? WOW. How did you that, obviously, not with a meter stick. On the surface, I don't believe measurement such as 114.602cm

Now you calculation of energy: I agree that is the change of PE.

so if you compare the final KE (1/2 mv^2) it compares well with the starting PE.