In: Physics
Friction Experiment
Procedure:
Use additional weight and put them on the hanger in order to achieve a constant velocity
Data A:
M1 |
M2 |
N |
Fnet |
mu |
.17 N |
.05 N |
1.7 N |
0.5 |
0.30 |
.22 N |
.055 N |
2.2 N |
0.55 |
0.25 |
.27 N |
.07 N |
2.7 N |
0.7 |
0.26 |
.32 N |
.07 N |
3.2 N |
0.7 |
0.22 |
.37 N |
.09 N |
3.7 N |
0.9 |
0.24 |
Procedure:
Formulas A:
Data B:
Distance |
Time |
A(experiment) |
MI(N) |
M2 |
A(theory) |
%Difference |
.5 m |
.93 s |
1.2 m/s^2 |
.12 N |
.1 N |
3.18 m/s^2 |
62.3% |
.5 m |
.64 s |
.61 m/s^2 |
.17 N |
.1 N |
2.13 m/s^2 |
71.8% |
Formulas B:
Question: I'm looking for the formulas worked out for both data tables
ex: plugging in what numbers into what equation to get answers from both data tables?
The diagram of experiment is approximately as shown below.
A)
There are two forces acting on the string. In experiment 1, a constant velocity is achieved so that the acceleration is zero. So, the net force is zero.
Fg is the force due to gravity on the mass M2.(Fnet in the table 1)
(Here, acceleration due to gravity is taken as 10 m/s2
The frictional force is given by
is the co-efficient of dynamic friction and N is the normal reaction force due to the two masses given by
Equating both sides, (since net force is zero)
B)
For the table 2, the net force is not zero, so that there is acceleration.
Theoritically,
Using Newton's second law, F=ma,
Take g = 10 m/s2 and you will get a(theory)
Experimentally, If distance traveled is d in time t
u, the initial velocity is zero,
Finally, percentage error is given by