Ch 6.1 Help
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Ch 6.1 Help
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 expression(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 1214)? 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:)
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 expression(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 1214)? 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 Guest
Re: Ch 6.1 Help
Question 1
(The answer is positive)
This is pretty straight forward. Use the expression:
W = 4.592N * 15.83
W = 72.688J
Question 2
(The answer is negative)
This is pretty straight forward. Use the expression:
W = 68.416J
Question 3
(The answer is negative)
Another straight forward question. Using
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 = 328.796J
(The answer is positive)
This is pretty straight forward. Use the expression:
W = (7.915N * sin(35.46)) * 15.83mW=(F*sin(Θ))*x
W = 4.592N * 15.83
W = 72.688J
Question 2
(The answer is negative)
This is pretty straight forward. Use the expression:
W = 3.69kg * 9.81m/s^{2} * 1.89mW=m*g*h
W = 68.416J
Question 3
(The answer is negative)
Another straight forward question. Using
We transform it to find v, soKE=(1/2)*m*v^{2}
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 = 7.981kg*9.81m/s^{2}*3.05mPE=m*g*h
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)
Guest01 Posts : 133
Join date : 20080919
Question #2 is WRONG
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 Guest
Re: Ch 6.1 Help
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.
Guest01 Posts : 133
Join date : 20080919
question 1
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?
guest05 Guest
Re: Ch 6.1 Help
If you look at 3' in the notes it says do not learn 6.1 by heart, watch out where the angle is sitting.
sbguest Guest
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