course PHY121
I emailed you a copy of the graphs and write up of video experiment 12.
Your work here is OK, per the email message I just sent you. No more lab work is necessary.
Table Slop
0.03925 0.1764
0.0332 0.14112
0.0257 0.10584
0.0201 0.07056
0.01262 0.03528
Effective Slope Force
0.0293 0.1764
0.0232 0.14112
0.0157 0.10584
0.0101 0.07056
0.0026 0.03528
We can determine that the straight line indicates that the a force vs. effective slope proportionality, because the line passes through the origin and through the data points.
We can determine that the slope of the line as .13-.06 / .02-.01 = .07/.01 = 7.
The weight of the cart is .550 kg * 9.8 m/s^2 =5.4 Newtons.
The weight of the car is equal to the slope of the graph within experimental error. We can conclude that for small slopes, the force required to restrain the cart from accelerating down the incline is equal to the weight of the cart times the the effective slope of the incline and the net force accelerating a cart coasting down that slope is equal to the weight of the cart times the effective slope of the incline.
The acceleration of a low friction cart down a small slope is equal to the accelerration if gravity times the effective slope.