Name ______School ______Date ______
AP/Honors Lab 6.1 – Conservation of Mechanical Energy; Work
Purpose
To study conservation of Mechanical Energy for a cart moving along an incline
To observe the scalar nature of energy
To examine the non-conservative nature of the friction force.
Equipment
Virtual low friction cart and track with pulleys, strings, and masses from the Dynamics Lab
PENCIL
Explore the Apparatus
Open the Virtual Dynamics Lab on the website. You should be familiar with most of the features of ramp and cart system from your study of dynamics. There are just a few new features which you should familiarize yourself with now.
Setting the initial velocity
The control panel below the right end of the track is used to control to control the initial velocity of the cart as well as the static and kinetic friction between the cart and track. Its appearance changes according to the situation. The figures below shows the three ways it can look when used to set the initial velocity.
Figure 1a shows the initial view of the initial velocity control. If you click [Go at known Vo] the cart will instantaneously acquire a velocity of +100 cm/s. Vo can range from-300 cm/s to +300 cm/s. You can change the direction of the velocity by inserting a – sign by hand or use of the Vo stepper gadget ( ) to move through the range of ± velocities. /
Figure 1a
In figure 1b the numeric stepper for Vo? has been adjusted to setting #3. The Go button changes to “Go at unknown Vo” to indicate that the cart will acquire some unknown (to you) velocity if the button is pushed. A setting of #3 will always produce the same Vo.
The clock icon at the upper right allows for a 3-second pause between when you press the Go! button and when the cart actually starts to move. It also automatically turns off the brake if it is set. This allows you to launch a cart from rest on an incline. /
Figure 1b
Adjusting the static and kinetic friction
Figure 2 shows the similar controls for static and kinetic friction. The two left steppers show the default values for μKinetic and μStatic. These steppers allow you to adjust these two values independently. Well, almost independently. μK must remain a respectful distance below μS. μKS? works similarly to Vo? That stepper allows you to choose from among several pairs of fixed but unknown (to you) μK and μS values. /Figure 2
Calculating Δh for potential energy calculations
In order to calculate the change in potential energy you’ll need to find the change in height of the center of gravity of the cart between two points. In Figure 3 you see the cart at two positions, original and final. Of course these would be reversed if the cart were going down the ramp. The ramp angle θ can be read from the protractor attached to the track. And the positions xo, and xf can be measured using the ruler. You may also be using photogates when you make these measurements in which case the ruler can just measure the positions of the two photogates.You can then find Δh = hf – ho using /
Figure 3
I. Conservation of Mechanical Energy
The kinetic energy of a body is the energy associated with its motion.
Kinetic Energy,
The potential energy of a physical system is the energy associated with the arrangement of the objects making up the system. This energy is due to the forces among the objects in the system and the work that those forces can do in changing the arrangement of the objects in the system.
In the case of gravitational potential energy for an object near Earth, the energy is due to the force of gravity acting between the objects.
If an object at some height above the earth is allowed to fall to a lower altitude, the loss in potential energy of the system of the earth and the object would be equal to the work done by the force of gravity. In such a case the motion of the earth is negligible and we usually say that it is the falling object that loses the energy.
The change in energy of the object is given by
where h is measured relative to some arbitrary level where we set h = 0.
We define the sum of the kinetic and potential energy of an object (or system) as its total mechanical energy. That is,
ME = PE + KE
A swinging pendulum is a good example of a system which undergoes a steady change in PE and KE. As the bob rises, its PE increases and its KE decreases. As it swings back downward the reverse occurs. If there are no forces other than gravity acting, the decrease in one type of energy is exactly matched by the increase in the other. Thus the total ME remains constant.
If the work on the object by non-conservative forces such as friction is zero, the total mechanical energy will remain constant. In that case we say that mechanical energy is conserved.
If Wnc = 0,
PEo + KEo = PEf + KEf
or Principle of Conservation of Mechanical Energy
Δ PE + Δ KE = 0
A car moving up or down a hill is another example of energy being converted between KE and PE. For a real car the total mechanical energy would not remain constant since there are other forces involved.
Ia. Δh for a cart traveling up a frictionless ramp and coming to a halt.
Consider a cart traveling up a frictionless ramp. It will start with a certain initial velocity and come to a halt after rising by a certain Δh. Your goal will be to predict and verify experimentally its change in height. To do this using kinematics and dynamics, you’d first need to find the acceleration using F = ma, and then use the acceleration in a kinematics equation. You’d also have to take into account the vector nature of the forces and motion.
With mechanical energy you need only look at the initial and final states (energy) of the system. And because energy is a scalar quantity, there are no vectors involved.
For the cart, the change in PE, ΔPE = mg(hf – ho) = mgΔh. You should be able to determine Δh from Figure 3 and trig.
The initial kinetic energy, KEo, can be found from the empty cart’s mass and its initial speed which will be set using the initial velocity control discussed above.
1. Set the ramp angle to +3°.
2. Set recoil to 0 and leave it there for all parts of the lab.
3. Set vo = 1.30 m/s. (130 cm/s) (Note: vo is not zero. The cart starts at 1.30 m/s)
4. KEo = ______J for the 250-g cart at 1.30 m/s. (Careful with units.)
5. Turn on the ruler. Use right-click + “Zoom in” and “100%” as needed for making precise position adjustments.
6. With the brake on and the cart mast at x = 20.0 cm, click [Go at known Vo] to start the cart.
7. Vf = ______m/s at the top of the cart’s travel.
8. KEf = ______J
9. Using ΔPE + ΔKE = 0, calculate the cart’s predicted change in height, Δh.
Δ h = ______m
Show calculations here.
Let’s see how accurate this prediction is. You can’t directly measure Δh with this apparatus. In order to determine the experimental value for Δh, you need to know xo, and xf. (See Fig. 3)
10. Turn on the motion sensor and again launch the cart from x = 20.0 cm.
11. Turn off the sensor when the cart returns to its starting point.
Look through the position data in the right column of the data table. Xo, and xf are the initial and maximum positions of the cart going up the track. How can you find xo, and xf from this data? Be careful, the data includes the motion back down the track which is not what you’re looking for.
xo = ______m xf = ______m
12. Calculate Δh from xo, and xf and the ramp angle as explained with Figure 3.
Δh = ______m
Show calculations here.
Ib. The relationship between speed and kinetic energy.
Here’s your chance to get a better understanding of the non-linear relationship between speed and kinetic energy. This is something that gives many students trouble.
Question: If you wanted the cart to go half as far up the ramp, that is, Δh/2, that means half the ΔPE, which would require just half the initial KE. You used Vo = 1.30 m/s before. What Vo would you use for Δh/2?
Find it experimentally first. Turn on the ruler. Add three photogates to the apparatus just for reference markers. (Click the radio button beside Photogates, then reduce # Gates to 3.
Put the first gate at 20 cm, the third one at the xf(orig) you found in part 1a (which is also at the original hf(orig)). You’ll put the second, middle one at the new xf(half) (hf(half)). The new xf(half) should be half-way between xo, and the original xf(orig).
1. New xf(half) = xo + (xf(orig) – xo)/2 = ______m (Think about it.)
You now have a target to shoot at. You first need to experimentally determine the proper Vo to give the cart ½ its previous KE. How about vo/2 = .65 m/s? Try it.
2. How’d that work out? Too fast or too slow? (Circle one.)
Here’s the problem and how to think about it.
KEo = ½ mvo2. So KEo is directly proportional to vo2, not vo. So vo/2 doesn’t give you KEo/2
So halving vo doesn’t halve KEo, halving vo2 does!
So if vo = 1.30 m/s
vo(orig)2 = _____ m2/s2
vo(new)2 = = _____ m2/s2
vo(new) = _____ m/s
The proof is in the testing. Give it a try and see if you need further head-scratching.
Because gravity was the only force acting with a component along the direction of motion, mechanical energy was conserved. That’s why we call gravity a conservative force. The same happens as the cart rolls back down the ramp.
3. If the cart is launched at 1.30 m/s up the ramp, at what speed should it be traveling when it returns to the bottom?
m/s
Similar to gravity, a spring can store up and release energy when it’s compressed or stretched. A perfect spring would act to exert conservative forces. Describe the motion that we’d see if we had a perfectly springy recoil bumper at the bottom of the track.
Give it a try with Vo = 1.30 m/s. For best results turn the recoil value down to zero. Then launch the cart. Then quickly restore the recoil value to 1. Not bad. (It sometimes fails and goes increasingly higher.)
II. Work by friction
Of course, most of the time mechanical energy is not conserved. The engine of a car causes the wheels to push backward on the road and the road acts by pushing the car forward. This forward push acts to increase the mechanical energy of the car. Likewise, air and rolling resistance works against the motion of the car to rob it of mechanical energy.
These two external forces that do positive or negative work to the car are called non-conservative forces. A force like a tail wind that acts to increase the energy of the system does positive work. Likewise, a force like friction that removes energy from a system does negative work. In either case we can say
Wnc = ΔPE + ΔKEo
where Wnc is the total work done by all non-conservative forces acting. Wnc can be either positive or negative.
In this part of the lab we’ll look at the negative work done by friction. We’ll use a level track and observe the decrease in velocity of the cart when friction is acting. Figure 4 shows the setup.
Figure 4
A cart traveling initially at some velocity, vo, at x = 10 cm, slows steadily as it moves from left to right. At x1, it has a velocity, v1. At x2, it has slowed to v2. It is slowed by a kinetic friction force with some friction coefficient, μk. The velocities, v1, and v2, can be found from the width of the purple card and the time, dT, for the card to pass through each photogate. These are actually average velocities, but the average velocity is approximately equal to the instantaneous velocity at the mid-point.
The work done by friction is given by
where
Note that the work is negative since the friction force acts opposite the direction of the motion. v22-v12 is negative as well.
Your task is to determine the value of μk experimentally and compare it to its known value.
- Set up two photogates as shown above at x1 = 40 cm, and x2 = 160 cm.
- Set the card width to 10 cm.
- Position the cart all the way to the left bumper and set its known Vo to 200 cm/s.
- Turn on the friction pad and set known value for μk to .10.
- [Take Gate Data] (Click the Take Gate Date button, etc. for next two)
- [Go at known Vo]
- [Stop Gate Data] after the cart stops moving.
Occasionally you’ll find that one of the dT values is negative. If this happens, just try again.
- Record your data.
Δx = ______m (distance between gates)
μk = ______(μk is the kinetic friction coefficient, no units)
mc = ______kg (mc: the mass of the cart, in kg)
dT1 = ______s (time to pass through first gate)
dT2 = ______s (time to pass through second gate)
v1 = ______m/s (average speed through first gate)
v2 = ______m/s (average speed through second gate)