# Ch 6.1 Help

## 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 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:)

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 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**- Guest

## Re: Ch 6.1 Help

*Question 1*(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/sW=m*g*h

^{2}* 1.89m

W = 68.416J

*Question 3*(The answer is negative)

Another straight forward question. Using

We transform it to findKE=(1/2)*m*v^{2}

**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 = 7.981kg*9.81m/sPE=m*g*h

^{2}*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)

**Guest01**- Posts : 133

Join date : 2008-09-19

## 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 : 2008-09-19

## 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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