Urgent help needed with F=ma concept! Where did I go wrong? Help please

  • #1
Urgent help needed with F=ma concept! Where did I go wrong?? Help please

Use the exact values you enter to make later calculations.

A group of students performed the same "Newton's Second Law" experiment that you did in class. For this lab, assume g = 9.81 m/s2. They obtained the following results:

m1(kg) t1(s) v1(m/s) t2(s) v2(m/s)
0.050 1.2000 0.2500 1.7862 0.4123
0.100 1.2300 0.3240 1.6364 0.6502
0.150 1.1500 0.3820 1.4809 0.8067
0.200 1.1100 0.4240 1.3988 0.9511


where m1 is the value of the hanging mass (including the mass of the hanger), v1 is the average velocity and t1 is the time at which v1 is the instantaneous velocity for the first photogate, and v2 is the average velocity and t2 is the time at which v2 is the instantaneous velocity for the second photogate.

(a) Use Excel to construct a spreadsheet to do the following. (You will not submit this spreadsheet. However, the results will be needed later in this problem.)

(i) Enter the above data.

(ii) Compute the acceleration, a, for each trial.

(iii) Create a graph of the hanging weight m1g vs. the acceleration. Hint

This is what I did
Excel1_zpsa2682c14.png


(iv) Use the trendline option to draw the best fit line for the above data and determine the slope and y-intercept from it.

And then I did,
Excel2_zps47b72cf9.png


(v) report your results below.

slope = __1.045__ kg
markSprite.png
Partial credit: Your answer received partial credit.
y-intercept = _0.2345_ N
markSprite.png
Partial credit: Your answer received partial credit.

Whatttt?????????? What am I doing wrong???????

(b) Use the information you obtained from your graph to determine the total mass of the system M = m1 + m2.
M = _1.045 kg_ Partial credit: Your answer received partial credit.

(c) Using the information you obtained in parts (a) and (b), predict what the value of the acceleration would be if the value of the hanging mass were increased to m1 = 0.40 kg.

a = ______ m/s2
 

Answers and Replies

  • #2
834
2


You graphed force on the x-axis and acceleration on the y-axis. How can the y-intercept be in newtons, then?
 
  • #3


You graphed force on the x-axis and acceleration on the y-axis. How can the y-intercept be in newtons, then?
Ohhh ok...

So I switched it and got
slope - 0.956 kg
y-intercept - 0.2248 N


b. 0.956 kg
c. a = ?

How would I solve for c?
 
  • #4
834
2


The equation of the line gives you the relationship between acceleration and force for this system. If you know one, you should be able to algebraically solve for the other.
 
  • #5


The equation of the line gives you the relationship between acceleration and force for this system. If you know one, you should be able to algebraically solve for the other.
y = 0.946a + 0.2248

ma = 0.946a + 0.2248
(0.40)a = 0.946a + 0.2248
???
 
  • #6


The equation of the line gives you the relationship between acceleration and force for this system. If you know one, you should be able to algebraically solve for the other.
I need help :(
 

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