Question

In: Physics

Friction Experiment Procedure: On the pulley place 50 grams weight on the block which is 0.05kg...

Friction Experiment

Procedure:

  1. On the pulley place 50 grams weight on the block which is 0.05kg
  2. Attach the wood block using a string and hanger
  3. Place the hanger on the other side of the pulley

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:

  1. Place 50 grams on top of the wood block and 50 grams on the hanger and observe acceleration. Use additional weight if necessary
  2. Record the time for the motion
  3. Calculate experimental acceleration
  4. Calculate theoretical acceleration
  5. Compare using percent error formula
  6. Repeat one more time

Formulas A:

  • M1= Mw+n*50 (n is an integer, start with n=1, go to n=4)
  • N=M1 g
  • Fnet= M2 g
  • Mu= Fnet/N

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:

  • A(experiment)= 2d/t^2
  • Average mu= 0.25 --> mu=0.25 for equation below
  • A(theory)= (M2 g- mu M1 g)/(M1+M2)

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?

Solutions

Expert Solution

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


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