please can someone solve for me part 3 just please i have for part 2 case 1: A(m
ID: 1585299 • Letter: P
Question
please can someone solve for me part 3 just please
i have for part 2
case 1:
A(m)= 0.501m
B(s) = 3.43 ms
B(m) = 0.098 m
case 2:
A(m)= 0.501
C(m) = 0.43m
case 3:
C to B to A (s): 3.58 ms
A to B to C (s): 3.80 ms
C to B to A (m/s): 2.79 m/s
A to B to C (m/s): 2.63 m/s
mass of cart (kg): 92.8 g
This "U" shaped piece will be used, with the photogate sensor, to measure the speed of the cart at the position of the sensor. It works as follows: When the infrared beam across the photogate is broken by the leading edge of the "U" (labeled 1 in the figure), an electrical signal is generated A second signal is generated when the beam is broken a second time by the edge of the "U" (labeled 2) that crosses the beam just after the gap. The signal acquisition and control unit will display the time (in milliseconds) between these two signals. The spatial separation between edges 1 and 2 of the "U" is 1.0 cm. This information allows you to calculate the speed of the cart at the position of the photogate The signal acquisition unit should already be powered on and ready for use. Ask for help in the use of this unit as needed Part 2: Each table will perform three measurements for three different cases involving changes in kinetic and/or potential energy of the cart Position the cart at the top of the left side of the ramp (position A in Figure 1 above). Carefully measure the height ha (to the nearest mm) from the table surface to the "center of gravity" (CG) of the cart, which is indicated in Figure 2 Height hafrom the table to the CG of the cart at position A (m) = Gently tap the cart so that it rolls freely down the ramp from position A toward position B. After the cart crosses through the photogate, record the crossing time displayed on the meter, and calculate the speed (1 cm divided by the crossing time) Crossing time at B (s) Speed of cart at B (m/s) = Place the cart at rest at position B, move the motion sensor out of the way, and measure the height hs (to the nearest mm) from the table to the CG of the cart Height hB from the table to the CG of the cart at position B (m) = Case 2: AK =0. AU 0 Position the cart at the top of the left side of the ramp (position A in Figure 1). Again measure the height ha (to the nearest mm) Height ha at position A (m)- Gently tap the cart so that it rolls freely down the incline from A, through the flat section, and up the ramp on the right towards position C. At the instant that the cart reaches zero speed, grab hold of it and without altering its position, measure the height hc from the table to CG of the cartExplanation / Answer
Hello, As this is an experimental question, I can only explain how to do calculations.
Based on the given information, I think the the height at point A is 0.501 m and point B is 0.098 m. Therefore, change in potential energy .
Delta P.E = mghA -mghB = 0.0928*9.81 ( 0.501 - 0.098) = 0.3668 JOULES
similarly you can find change in kinetic energy,
Remember, you need to find velocity at point B which we can calculate by using
v = sqrt(2gh) where h is the height through which the cart falls from point A to B. However, this equation assumed the track to be frictionless. But I see that you have given time to fall from point A to point B. so it's better to use kinematics. use h = ut + 1/2at2 where h is height through which it falls, u = 0 (rest) t is the time, get the acceleration and then use v2 - u2 = 2ah to get v (final velocity at point B)
Note - kinetic energy at the top is zero as the cart is at rest.
After you have found out delta K.E just add delta K.E and delta P.E
The same procedure is valid for case 2 and case 3.
It's very simple to answer subsequent parts. you have calculated everything. Now just compare the delta K.E + delta P.E to just change in P.E and see how much energy is lost due to friction.
If you have any doubt, please let me know
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