Question

 

1. Record your lab data regarding photogate positions and times in the table below, converting the units as needed. Order the entries from shortest distance to longest.

#	Position (m)	Time (s)
1	40 cm	0.05 s
2	50 cm	0.05 s
3	60 cm	0.15 s
4	80 cm	0.25 s
5	90 cm	0.30 s
6	100 cm	0.35 s
7	110 cm	0.35 s
8	120 cm	0.40 s
9	130 cm	0.40 s
10	140 cm	0.40 s
11	160 cm	0.50 s
12	190 cm	0.55 s

Data Analysis

2. In this lab, you recorded the distance that a ball fell and the time it took for the ball to cover this distance. How can this information be used to find the acceleration due to gravity?

   suppose the distance traveled by the ball is s, and the time taken to fall is t,then its initial velocity is v=0. Using newton's second equation of motion, s=vt+ 1/2 gt2. so g= 2s/t2. by using this equation, we can find that the acceleration is due to gravity.

3. Calculate the distance Δythat the ball fell from the starting position to the photogate position and the square of the falling time Δt²for each entry in the table in question 1 and enter them into the table below. To calculate the distance, subtract the starting position (20 cm) from the photogate position.

#	Δy(m)	Δt2
1	20
2	30
3	40
4	60
5	70
6	80
7	90
8	100
9	110
10	120
11	130
12	170

4. Graph Δyvs. Δt²from the table in question 3 and use this graph to find g. To do so, use the graphing tool to plot the Δyvalues on the y-axis and the Δt²values on the x-axis. Add a linear fit line and save your graph to your portfolio. Find the slope of the line and relate it to gusing the relationship .

Conclusions

1. How well does your experimentally determined value of gagree with the theoretical value of 9.81? Calculate the percent error in your answer using the following relationship
   
2. The experimentally determined value for gis below the expected value of 9.81 m/s². Other than the precision of your measurements, is there any other factor that would cause your answer to be too low?

   Air friction can cause an experimentally determined gravity acceleration to be less than what it really is. For best precision, the drop needs to be in a vacuum. But your measurement seems to be some sort of error in measurement or calculations, assuming you are not too high above sea level on earth. Gravity also varies even then, dependent on where on earth you measure it.

3. An astronaut, who weighs 805 N on Earth, lands on a planet that has 3 times the mass and twice the radius of Earth. What is the astronaut’s weight on this new planet? Show your work.

   Wp=m3/6Ge= mge=mge x 3/4=3/6 x 805 Wp=603.75 N

NOTES:

-moved the photogates to 40 cm and 100 cm

the ball is released from 20 cm

top photogate time 0.05 secs

bottom photogate 0.35 secs

force plate timer 0.60 secs

-Photogates at 50 cm and 120 cm

released from 20 cm

top photogate time 0.05 sec

bottom photogate time 0.40 sec

force plate timer 0.60 sec

60 cm and 130 cm

top photogate time 0.15 sec

bottom photogate time 0.40 sec

force plate time 0.60 sec

80 cm and 140 cm

top photogate time 0.25 sec

bottom photogate time 0.4 sec

force plat etime 0.60 sec

90 cm and 160 cm

top photogate time 0.30 sec

bottom photogate 0.50 sec

force plate time 0.60 sec

110 cm and 190 cm

top photogate time 0.35 sec

bottom photogate time 0.55 sec

force plate time 0.60 sec

 

Answer 1