Are Ramp Length and Angle Key to Reducing Work in Physics Problems?

[scope]

Problematic Physics Problems (please help)

Hi, I recently received a couple of grade 11 physics problems that I'm having a tough time with. I found this forum, and I hope that I can get some hints on how to go about solving these problems. The problems are all regarding work/energy/power.

I partly solved each problem that I had trouble with, and I'll include my incomplete solutions below the problem. However, I don't know if I'm on the right track or not. I hope someone can help me on these.

1) If you want to load a box onto the back of a truck, you usually use a ramp that connects the ground to the back of the truck. One furniture mover claimed that less work can be done in loading the truck if the length of the ramp is increased in order to decrease the angle of the ramp with respect to the horizontal. Is his claim valid?

-Ok. What I tried for this problem is to substitute in numbers and try to see which situation would create less work required. The height of the truck is the same in both cases, and I imagined that to be 10m. One situation had a ramp angle of 70 degrees, and the other had an angle of 10 degrees. I then found the lengths of the hypotenuse in each situation (10.64m and 57.55m, respectively).
I know that work done equals the force parallel to motion multiplied by the distance travelled. Where do I go from here? Do I assume that the force applied on the box is the same in each case? If so, then more work would be required for the longer ramp. But I have a feeling that this is not correct.

2) A boy fires a 60g stone with a slingshot directly upwards. The pebble leaves the slingshot at 35m/s. How high above the ground will the stone reach if it is fired straight up? (Assume no friction). At what speed would an 80g pebble have to be fired to reach the same height as the first pebble? (This pebble is also fired straight up).

-Ok. I basically solved this problem, but I want to know if I'm correct.
I know that velocity2 of the pebble in both cases is 0, and velocity1 for the first pebble is 35m/s. With this info, I found the change in kinetic energy of the first pebble. The amount of kinetic energy lost by the pebble would equal to the amount of gravitational potential energy gained by the pebble. So now I used the formula Eg=mgh to calculate the maximum height of the pebble. I found this to be 62.4m.
I know that the second pebble that's 80g has to reach this same height. I used the formula Eg=mgh to find the change in gravitational potential energy. I then used this value to solve for velocity1. I found that the velocity needed to propel this 80g pebble is 35m/s. But this is the same velocity used to propel the 60g pebble! Logically, shouldn't the velocity needed to propel the heavier pebble be greater than the velocity needed to propel the lighter pebble? Both pebbles reach the same height, but one is just heavier...so should the velocity needed to propel the heavy pebble be greater than 35m/s? What am I doing wrong?

3)A 1000kg car travels up a hill that is 170m long. The hill makes an angle of 3.37 degrees with the ground. The car slows from a speed of 15m/s at the bottom of the hill to a speed of 12m/s at the top, and this journey took 13s. The force of friction was 600N. Find the average power of the car.

-Yikes, this problem is very difficult for me. I don't really know how to go about solving it. I first found the acceleration of the car, which was -0.2308m/s^2. I then found the change in kinetic energy and the change in gravitational potential energy. I know that I have to find the work done by the car, but I don't know how to do it. Does acceleration due to gravity play a role? Where do i subtract the work done by friction? I would appreciate some pointers / help with solving this problem.

4) Two masses are connected by a rope over a light and frictionless pulley. The mass on the left is 5.0kg and the mass on the right is 15.0kg. The 15kg mass is suspended 2.5m above the ground, while the 5kg mass is resting on the ground. The system is released so that the 15g mass moves downward and the 5kg mass on the other side is pulled upward. Determine the maximum height attained by the 5kg mass. Assume that the rope is long enough so that the 5kg mass will never hit the pulley.

-This one is also giving me a really hard time. I know it involves finding gravitational potential energy / kinetic energy and then substituting the energy of one mass into the other mass. I can find the kinetic energy gained by the heavier mass when it falls to the ground. Where do I go from here? How can I solve the problem? Since the 2 objects have different masses, I don't know how to manipulate the energies...Help will be appreciated.

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That's all...I hope someone can help me with these problems ASAP. Thanks a lot in advance.

 

[Mulder]

I'll give you a suggestion or two on the first two for now, I'm short of time but will help on 3 and 4 if no one has done later.

Originally posted by scope

1) If you want to load a box onto the back of a truck, you usually use a ramp that connects the ground to the back of the truck. One furniture mover claimed that less work can be done in loading the truck if the length of the ramp is increased in order to decrease the angle of the ramp with respect to the horizontal. Is his claim valid?

-Ok. What I tried for this problem is to substitute in numbers and try to see which situation would create less work required. The height of the truck is the same in both cases, and I imagined that to be 10m. One situation had a ramp angle of 70 degrees, and the other had an angle of 10 degrees. I then found the lengths of the hypotenuse in each situation (10.64m and 57.55m, respectively).
I know that work done equals the force parallel to motion multiplied by the distance travelled. Where do I go from here? Do I assume that the force applied on the box is the same in each case? If so, then more work would be required for the longer ramp. But I have a feeling that this is not correct.

OK work done is change in Energy, which here is all potential assuming it's pushed at constant 'v'. So it doesn't matter how long the ramp is made, if the object has to be pushed the same distance vertically each time then the Work done will be the same.

If you're thinking W = F.dx, you have to take 'dx' perpendicular to gravity, so you would have a 'cos(Q)' for ech different angle of elevation of the ramp, and so it would still work out that the work done for each angle was the same.

2) A boy fires a 60g stone with a slingshot directly upwards. The pebble leaves the slingshot at 35m/s. How high above the ground will the stone reach if it is fired straight up? (Assume no friction). At what speed would an 80g pebble have to be fired to reach the same height as the first pebble? (This pebble is also fired straight up).

-Ok. I basically solved this problem, but I want to know if I'm correct.
I know that velocity2 of the pebble in both cases is 0, and velocity1 for the first pebble is 35m/s. With this info, I found the change in kinetic energy of the first pebble. The amount of kinetic energy lost by the pebble would equal to the amount of gravitational potential energy gained by the pebble. So now I used the formula Eg=mgh to calculate the maximum height of the pebble. I found this to be 62.4m.
I know that the second pebble that's 80g has to reach this same height. I used the formula Eg=mgh to find the change in gravitational potential energy. I then used this value to solve for velocity1. I found that the velocity needed to propel this 80g pebble is 35m/s. But this is the same velocity used to propel the 60g pebble! Logically, shouldn't the velocity needed to propel the heavier pebble be greater than the velocity needed to propel the lighter pebble? Both pebbles reach the same height, but one is just heavier...so should the velocity needed to propel the heavy pebble be greater than 35m/s? What am I doing wrong?

No. 35 m/s however heavy the pebble. Remember v^2 + u^2 = 2as - the acceleration on an object - ie gravity is the same whatever the object's mass. Only the Force required to elevate a heavier object to the same height would increase.

 

[scope]

..

Hmm. Thanks a lot Mulder. I could still use a bit of explanation for #1. I kept trying to solve number 3 and 4, but I get stuck in the same place. Any help is most appreciated.

 

[Mulder]

OK, #1. You know that work done = change in Energy. Here, it's just the change in potential energy. You get that? Now the change in potential energy, mgh, will be the same regardless of the path you take to get it to that height! So whether you lift something straight up, or take a path with some horizontal component, it won't matter because the change in Energy only depends on the difference in height. That's the key point.

#3, hmm, is average power given my the driving force of the car x average velocity? You can work out the driving force resolving forces, ie. Reaction force and friction and using F = ma, since you can work out a, so I think that would be where to start on that one.

#4, looking at it, you're going to have firstly a time when the lighter mass moves up off the ground with tension in the rope until the heavier mass hits the floor. It's going to move 2.5 m in this part - you're given that. Then I believe the question wants you to work out how much further the lighter block will continue once the tension disappears - in effect in freefall, (except its going up and not down), so you'll just need to work out the initial velocity to find this distance - in other words the velocity that the heavier block is traveling at when it hits the ground. Can you get anywhere on this?

 

[GSXR750]

If you don't neglect the friction with the first problem, the amount of work needed increases with the length of the ramp.

If you call theta the angle of the ramp,

The friction force is F=mu*m*g*cos(theta)

and cos(theta)=sqrt(1-(h/l)^2) withh the heigth and l the length of the ramp.

The amount of work is: w=mu*M*g*sqrt(1-(h/l)^2)*l
and this increases with l.

So that statement is wrong.

 

[Mulder]

^Yes I should have mentioned that. I assumed we were discounting (what would admittedly be a large amount of) friction.

 

[scope]

Ok, thanks for all your help. I solved all the problems except number 3. Help appreciated.