|
|
|
|
Junior Member
|
|
Feb 18, 2011, 08:08 PM
|
|
work and energy
http://www.smartphysics.com/images/content/mechanics/ch9/blockonincline.png
A mass m = 12 kg is pulled along a horizontal floor with NO friction for a distance d =5.2 m. Then the mass is pulled up an incline that makes an angle θ = 35° with the horizontal and has a coefficient of kinetic friction μk = 0.34. The entire time the massless rope used to pull the block is pulled parallel to the incline at an angle of θ = 35° (thus on the incline it is parallel to the surface) and has a tension T =55 N. The speed of the block right before it begins to travel up the incline is 6.249 m/s.
How far up the incline does the block travel before coming to rest? (Measured along the incline.)
I got 5.17 m, but I am not sure if I am right! If I did wrong, please tell me how I should do it!
On the incline the net work done on the block is:
1. positive
2. negative
3. zero
it think it should be negative because the work done by tension is positive, but the magnitude of work done by friction and gravity is negative and is much bigger in magnitude, so the net work done should be negative, right?
|
|
|
Uber Member
|
|
Feb 19, 2011, 01:18 AM
|
|
Along the incline, the forces that apply are friction, weight and the pulling force.
taking g = 9.8 m/s^2
I get a = 4.37 m/s^2
So, distance up is:
I get
For the second part, remember that work done is equal to force x displacement. A force was applied and the mass moved a certain distance... so?
|
|
|
New Member
|
|
Oct 12, 2011, 11:56 AM
|
|
There is a simpler way to solve this problem without having to rely on the equations of motion. Remember:
1. W[net] = Force[net] * displacement
2. W[net] = delta K = (1/2)m*v^2 - (1/2)m*vi^2 where vi = initial velocity
Using equation 2, (1/2)m*v^2 = 0 since v^2 will be 0 once the object comes to rest.
Thus, W[net] = -(1/2)m*vi(2)
Substituting this into equation 1, we get:
W[net]/F[net] = displacement
where F[net] = F[kinetic friction] + F[weight] - F[Tension] = u[k]*g*Cos[theta] + m*g*Sin[theta]-T
==> displacement = -(1/2)m*vi^2 / ( u[k]*g*Cos[theta] + m*g*Sin[theta]-T )
You can plug in the numbers yourselves, the equation is now in terms of all known variables.
Having done the calculations, you will see that the answer matches that of the answer above me. This is just another solution, relying solely on the equations of Work.
|
|
|
Uber Member
|
|
Oct 13, 2011, 01:04 AM
|
|
Indeed, but I like to use both independently, to make sure my answers are correct.
I was actually planning to show the asker the other method too once (s)he gets to understand how to solve the problem, but since (s)he never got back :(
|
|
Question Tools |
Search this Question |
|
|
Add your answer here.
Check out some similar questions!
Work and Energy
[ 1 Answers ]
A man is pulling a boat through a narrow canal. If the tension on the rope is 1610 N and 138569 J of work is done in towing the boat 90 meters, find the angle between the rope and the forward direction.
Energy, Work and Power
[ 2 Answers ]
caculate the number of revolutions of the shaft required to raise a load through 1.2m.
Diameter of the shaft=10.0m
mass of the load=0.5kg
time taken to raise the load through 1.2m=5.0s
assuming that the thickness of the thread can be ignored and that the gravitational force acting on a mass of...
Kinetic Energy and Work
[ 3 Answers ]
A 25 kg box is initially at rest slides down a 2.0 m incline that is 15 degrees from the horizontal w/ a constant kinetic friction of 10N between the box and incline. What KE would the box have at the bottom of the incline?
I wanted to use the KE formula, but v is not given so I am stuck... can...
Work and Energy
[ 1 Answers ]
I have been trying to work on this Physics problem for hours now, and I'm not getting anywhere... a box has 500 J of work done on it by an applied force. How fast will it move after the work is done? That's not the exact question, but that's the idea. I know the general formula is W=(Fs) cos......
Work,kinetic energy
[ 1 Answers ]
In the equation for work why do you multiply force times displacement distance could'nt you just multiply forcec times the time it took the object to cover the displacement distance?
View more questions
Search
|