Ch 6.2 Help
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Ch 6.2 Help
Question 1
A person starting with an initial downward velocity of 2.446 m/s falls a vertical distance of 0.858 m. What is the final velocity of the person (see sheet 15)? Neglect friction in the air and indicate with a negative (positive) sign whether your equation for the conservation of the mechanical energy leads (does not lead) to equation (2.7) in Ch 2, that is an equation which does not use any energy principle.
How to solve
(Answer is negative)
v_{final}=√(v_{0}^{2}+(2*g*h))
v_{final}=√(2.446m/s^{2}^{2}+(2*9.81m/s^{2}*0.858m)
v_{final}=√(5.982916+16.83396)
v_{final}=√(22.816876)
v_{final}=
Question 2
A roller coaster, initially at rest, starts from 18.89 m above ground, travels down into a valley 3.073 m below the start point, rises back up a height of 1.975 meters and descends from there to the ground. With what speed does it arrive at the bottom (see sheet 1618)? Neglect any friction and indicate with a positive (negative) sign whether the roller coaster after having fallen from the initial height straight down to the ground would have arrived with the same (a different) speed as calculated at the bottom (assuming no friction, see sheet 18').
How to solve
(Answer is positive)
We can ignore the rise and fall of the roller coaster so our formula becomes
v=√(370.6218)
v=
Question 3
A 1.778 kg object slides on a horizontal plane and is stopped by friction after a distance of 1.1027 m. What is the initial kinetic energy when the coefficient of friction of the objectplane interface is 0.1627 (see sheet 15,19,19')? Indicate with a negative (positive) sign whether the object changes (does not change) its potential energy during the slide.
How to solve
(Answer is positive)
KE=F_{fr}*m*g*h
KE=0.1627*1.778*9.81*1.1027
KE=
Question 4
A block, starting at rest and from a height of 0.2676 m, slides down a frictionless inclined plane. At the bottom the block continues to slide horizontally and is brought to a halt after 1.0784 m by friction. What is the coefficient of friction of the blockhorizontalplane interface (see sheet 15,17,1819')? Indicate with a negative (positive) sign whether the block changes both its kinetic and potential (only its kinetic) energy during the slide on the inclined plane.
How to solve
(Answer is negative)
F_{fr}=h/x
F_{fr}=0.2676/1.0784
F_{fr}=
A person starting with an initial downward velocity of 2.446 m/s falls a vertical distance of 0.858 m. What is the final velocity of the person (see sheet 15)? Neglect friction in the air and indicate with a negative (positive) sign whether your equation for the conservation of the mechanical energy leads (does not lead) to equation (2.7) in Ch 2, that is an equation which does not use any energy principle.
How to solve
(Answer is negative)
v_{final}=√(v_{0}^{2}+(2*g*h))
v_{final}=√(2.446m/s^{2}^{2}+(2*9.81m/s^{2}*0.858m)
v_{final}=√(5.982916+16.83396)
v_{final}=√(22.816876)
v_{final}=
 Spoiler:
 4.777
Question 2
A roller coaster, initially at rest, starts from 18.89 m above ground, travels down into a valley 3.073 m below the start point, rises back up a height of 1.975 meters and descends from there to the ground. With what speed does it arrive at the bottom (see sheet 1618)? Neglect any friction and indicate with a positive (negative) sign whether the roller coaster after having fallen from the initial height straight down to the ground would have arrived with the same (a different) speed as calculated at the bottom (assuming no friction, see sheet 18').
How to solve
(Answer is positive)
We can ignore the rise and fall of the roller coaster so our formula becomes
v=√(2*9.81*18.89)v=√(2*g*h)
v=√(370.6218)
v=
 Spoiler:
 19.252
Question 3
A 1.778 kg object slides on a horizontal plane and is stopped by friction after a distance of 1.1027 m. What is the initial kinetic energy when the coefficient of friction of the objectplane interface is 0.1627 (see sheet 15,19,19')? Indicate with a negative (positive) sign whether the object changes (does not change) its potential energy during the slide.
How to solve
(Answer is positive)
KE=F_{fr}*m*g*h
KE=0.1627*1.778*9.81*1.1027
KE=
 Spoiler:
 3.129
Question 4
A block, starting at rest and from a height of 0.2676 m, slides down a frictionless inclined plane. At the bottom the block continues to slide horizontally and is brought to a halt after 1.0784 m by friction. What is the coefficient of friction of the blockhorizontalplane interface (see sheet 15,17,1819')? Indicate with a negative (positive) sign whether the block changes both its kinetic and potential (only its kinetic) energy during the slide on the inclined plane.
How to solve
(Answer is negative)
F_{fr}=h/x
F_{fr}=0.2676/1.0784
F_{fr}=
 Spoiler:
 0.2481
Guest01 Posts : 133
Join date : 20080919
question 2
Part one: With what speed does it arrive at the bottom? Which bottom does he mean; the valley or at the end of the ride?
Part two: Why do we neglect all of the heights given? In lecture it was impressed upon the class that it was all about the change in height.
Part two: Why do we neglect all of the heights given? In lecture it was impressed upon the class that it was all about the change in height.
super Mo Guest
ques 3
How did you find that equation, meaning did you manipulate a previous formula?
ihatephysics2 Posts : 1
Join date : 20081019
Re: Ch 6.2 Help
The bottom is the end and not the valley.
I believe the heights are neglected because it from point A to point B. If you go from point A at 10 mph to point B and then back to point A at 7 mph your velocity is zero because you start and stop at the same point. If you are driving on a hilly road, going up you may be a little slower and going down you may be a little faster, you can speed, stop at lights, do what you gotta so, however when you reach the end the velocity is only in respect to begging and end.
Which formula do you mean. I am pretty certain that the formula has been changed though.
I believe the heights are neglected because it from point A to point B. If you go from point A at 10 mph to point B and then back to point A at 7 mph your velocity is zero because you start and stop at the same point. If you are driving on a hilly road, going up you may be a little slower and going down you may be a little faster, you can speed, stop at lights, do what you gotta so, however when you reach the end the velocity is only in respect to begging and end.
Which formula do you mean. I am pretty certain that the formula has been changed though.
Guest01 Posts : 133
Join date : 20080919
Question 4
Question 4
by S Today at 5:38 pm
Could you please explain how you arrived at the equation for question 4? I think I understand that the work done from the top of the inclined plane, to the bottom, is zero (right?), but then I didn't know how to formulate the equation to get the answer.
Thanks.
by S Today at 5:38 pm
Could you please explain how you arrived at the equation for question 4? I think I understand that the work done from the top of the inclined plane, to the bottom, is zero (right?), but then I didn't know how to formulate the equation to get the answer.
Thanks.
S Guest
Re: Ch 6.2 Help
I actually believe I took the formula from Sheet 19'. I think I just understood it as this height has some energy, and if there is some friction acting on it that will give the coefficient of friction.
Guest01 Posts : 133
Join date : 20080919
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