Ch 6.1 Help

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Ch 6.1 Help

Post  gguest on Tue Sep 30, 2008 8:56 pm

Question 1: (1 point)

A box on a frictionless horizontal plane is pulled a distance of 15.83 m by an attached rope which makes an 35.46 degrees angle with the vertical and is under a 7.915 N tension. What is the work done (see sheet 3). Indicate with a negative (positive) sign whether the component of the tension perpendicular to the displacement is doing work (does not do any work).


(How to solve:)

Question 2: (1 point)

An object with a 3.69 kg mass is lifted by a force acting vertically upward a distance of 1.89 m. During the lift the object rises with constant speed. What is the work done by the upward force (see sheet 5)? Indicate with a negative (positive) sign whether the work done by the net force is zero (is not zero).


(How to solve:)


Question 3: (1 point)

A horizontal force of 17.98 N pushes a 1.775 kg block which is initially at rest on a horizontal frictionless plane for a distance of 0.8209 m. What is the final velocity of the block (see sheet 3,8,9)? Indicate with a negative (positive) sign whether you can (cannot) arrive at the same result using Newton's Law II and the expres​sion(2.7) from Ch 2, that is by not using any energy principle.


(How to solve:)


Question 4: (1 point)

A 7.981 kg box, initially on a table, is lifted a height of 3.05 m. What is the change in potential energy (see sheet 12-14)? Indicate with a positive (negative) sign whether the box at its final height has the same (different) potential energies relative to the floor and the table top.

(How to solve:)

gguest
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Re: Ch 6.1 Help

Post  Guest01 on Wed Oct 01, 2008 4:52 am

Question 1
(The answer is positive)

This is pretty straight forward. Use the expression:
W=(F*sin(Θ))*x
W = (7.915N * sin(35.46)) * 15.83m
W = 4.592N * 15.83
W = 72.688J

Question 2
(The answer is negative)

This is pretty straight forward. Use the expression:
W=m*g*h
W = 3.69kg * 9.81m/s2 * 1.89m
W = 68.416J

Question 3
(The answer is negative)
Another straight forward question. Using
KE=(1/2)*m*v2
We transform it to find v, so
v=√((2*KE)/m)

First we find KE for this formula. Note this isn't the actual KE formula, but when entered into the whole equation it works (make sense?)
KE = N*x
KE = 17.98N * 0.8209m
KE = 14.760N*m

Now plug in:
v = √((2*14.760N*m))/1.775kg)
v = √(29.520N*m/1.775kg)
v = √(16.631)
v = 4.078

Question 4
(The answer is negative)

Suddenly you realize that this homework was not hard at all and you can now go enjoy a soda.
PE is nearly the same formula as Question 1 and Question 2. Actually Question 2 and Question 4 are the same.
PE=m*g*h
PE = 7.981kg*9.81m/s2*3.05m
PE = 328.796J


Last edited by Guest01 on Sat Oct 04, 2008 12:58 am; edited 1 time in total (Reason for editing : Calculation errors)

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Question #2 is WRONG

Post  PHYSTUDE on Fri Oct 03, 2008 10:49 pm

why are you using the distance (x) from question #1, to answer number 2?? I keep using your solution and it marks it wrong. Then I used the height given in question 2, but thats not working either...can someone explain this to me? Thanks so much.

PHYSTUDE
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Re: Ch 6.1 Help

Post  Guest01 on Sat Oct 04, 2008 1:02 am

Sorry, I must have switched a few numbers and units as I was posting this with my own homework/quiz. I went back and corrected it. I used Newtons when it should have been kg, and the height wasn't the same, but I changed it now. If you only tried with the numbers given that's why it was wrong, you should now be able to get the same answer.

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question 1

Post  guest05 on Sat Oct 04, 2008 11:10 am

hey. in question 1 you used sin, but in the notes of chapter 6 (sheet 3) they are using cos. Im confused. Which one is right?

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Re: Ch 6.1 Help

Post  sbguest on Sat Oct 04, 2008 6:05 pm

If you look at 3' in the notes it says do not learn 6.1 by heart, watch out where the angle is sitting.

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