We are using the KET Virtual Lab. So far the only friction force we’ve encounter
ID: 1332863 • Letter: W
Question
We are using the KET Virtual Lab.
So far the only friction force we’ve encountered is the one supplied by the cart brake. It’s much too strong a force for our purposes. Instead we’ll use a friction pad like the brake. But this one is adjustable. In Figure 12 you see a snapshot of the cart traveling to the right at velocity Vo. A friction force, the net force, acts to the left, thus slowing the cart down.
In Figure 13 you see the wheels/friction pad control. When “wheels” is selected (Fig. 13a) the Vo controls are active. By selecting “friction pad” (Fig. 13b) the controls change to allow you to adjust k, and s. (They can’t both be selected at once. Each is shown in its selected mode.) We’ll use only kinetic friction which is controlled by the k stepper. We’ll start with it set just as it is in the figure. After one trial you’ll try to determine an unknown kinetic friction coefficient.
Click the friction pad radio button to turn on the friction controls. Set the friction controls as shown in Figure 13b.
Turn on the ruler. Move it above the dynamic vectors display.
Move the cart to x = 10.0 cm. (This is the position of the cart mast.)
Turn on the dynamic vectors: Check Velocity, Net Force, Friction
Click “wheels” and set Vo to 200 cm/s. Then switch back to “friction pad.Click Go
The cart should launch to the right and slow to a stop before reaching the right bumper. If not, retry the preceding instructions.
The cart slows down due to the friction force acting on it. From Newton’s second law we know
F = ma
Ff = ma
Ff = -k FN = -k mg
-k mg = ma
k = -a/g Equation 2
So if we knew a, we could divide by g to determine k.
How can we find the acceleration, a? If we launch the cart along the track, letting it come to a halt we know
vo = 2.00 m/s
vf = 0.00 m/s
x can be found with the ruler
a can be found from a single kinematics equation. You figure that part out.
Got it? Take the data you need and record it in Table 2. If your value for k is not close to .12, then you’ve done something wrong with your technique or your calculations. You should get a negative value for the acceleration.
Table 2 Determination of known k = .12
g = 9.80 m/s2
Show calculations below for a and k
vo = m/s
x1 = m
x2 = m
x = m
a = m/s2
k(experimental) = (no units)
Let’s try an unknown friction coefficient. Change the ks? value to 2 using the numeric stepper. You’ll quickly find that your 2.00 m/s is not a good choice this time. You can change it to whatever you like. But letting the cart go most of the way along the track will give better results than a short run. Note also that the new k should be much smaller than the previous value.
Table 3 Determination of unknown ks?
g = 9.80 m/s2
Show calculations below for a and k
vo = m/s
x1 = m
x2 = m
x = m
a = m/s2
k(experimental) = (no units)
The main question is:
What value did you measure for the unknown coefficent of kinetic friction?
Table 2 Determination of known k = .12
g = 9.80 m/s2
Show calculations below for a and k
vo = m/s
x1 = m
x2 = m
x = m
a = m/s2
k(experimental) = (no units)
Explanation / Answer
Hi,
I have gone through the question very clearly. The answer certainly lies on the lab experiments. Theoretically we can discuss here. The first task is for a fixed value of uk (0.12 shown in figure 13) in friction pad, calculate the acceleration of cart if the initial (wheels speed, Figure 13) and final velocity (zero since it comes rest), the distance travelled by the cart (can be seen the ruler Figure 12) is known.
Acceleration is measured using v2-u2 = 2 * a * dx
And also estimate k = -a/g whose value should be close to (0.12 shown in figure 13).
Similarly, now the change the value of ks in Figure 13 and similarly repeat the same calculations to estimate both a and k. Be cautious in choosing the values for ks and initial velocity
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