Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

Refer to http://bit.ly/2uCD7KV Lab handout for answers NEWTON’S SECOND LAW Prela

ID: 1632037 • Letter: R

Question

Refer to http://bit.ly/2uCD7KV Lab handout for answers

NEWTON’S SECOND LAW Prelab Quiz

QUESTION 1

How should the acceleration be found in each trial?

a. Fit the position vs. time graph to a parabola and multiple the coefficient in front of t^2 by 2.

b. Take the reciprocal of the slope of best-fit-line of the velocity vs. time graph.

c. Take the slope of best-fit-line of the velocity vs. time graph.

d. Taking the average of the acceleration vs. time graph.

e. Fit the position vs. time graph to a parabola and dividing the coefficient in front of t^2 by 2.


QUESTION 2

How do you find the frictional force from the tension vs. acceleration graph?

a. It's equal to the slop of the best-fit line.

b. It's equal to the y-intercept of the best-fit line.

c. It's equal to the reciprocal of the slope of the best-fit line.

d. It's equal to the x-intercept of the best-fit line.

QUESTION 3

Which is correct regarding the tension in the cord?

a. T < m(H)g

b. T = m(H)g

c. T > m(H)g

QUESTION 4


Why is it important to press the tare button on the force sensor before each run?

a. The force data may be offset and not have a proper zero value

b. The sensor will get too hot and shut down

c. Who cares why it's important, I just follow instructions without thinking.

d. Capstone will not start recording until tare is pressed

QUESTION 5

How is the mass of the cart, m(c) , found from the tension vs. acceleration graph?

a. It's equal to the y-intercept of the best-fit line.

b. It's equal to the x-intercept of the best-fit line.

c. It's equal to the reciprocal of the slope of the best-fit line.

d. It's equal to the slope of the best-fit line.

Explanation / Answer

q1) option c as we are getting only velocity vs time plots the best fit line slope would give the acceleration value neglecting the initial jerks if any.

q2) T- fr = mca , where fr is the frictional force

Thus, t= ma + fr

therefore, y intercept of the graph should give the frictional force

q3) mH g - T = mc a

therefore, mH g - mc a= T

this implies, mH g >T

q4) option a

q5) equal to slope of best fit line