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zell_D
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Hey guys, so after dealing with forces we started learning energy related problems. And I do get some problems but things get some what complicated. There are 3 questions that I need help on, the last one being the most difficult. One of them is conceptual that i wish to understand.
1. a 55 kg skier skis down a smooth frictionless ramp, slows down over a distance of 25 m on the flat portion at the bottom as shown in the diagram for number 1.
a) The skier pushes off at the top with a initial speed of 10 m/2. Find her speed at the top of the middle hump
b) The skier comes to rest at the end of the rough section. Find the average friction force experienced by the skier
2. You release a frictionless cart at the top of each of the 2 ramps. On ramp B, the cart is released from rest while the Ramp A cart is pushed with some initial velocity. The ramps have the same height as each other. The only difference is that one has a small bump up and the other a bump down. Which cart has the larger speed at the finish? why?
3. a) calculate the velocity of the ball when it just leaves the launching ramp (both magnitude and direction. be sure to specify a coord system.)
b) spring constant k = 800.0 N/m, the spring's compression is 3.00 cm, the mass of the ball is 50.0g, the height of the ramp is 15.0, and H (table's height) is 1.00 meter. With what total speed will the ball hit the floor? (use g = 10 m/s^2)
1. a) 0=[tex]\Delta[/tex]KE+[tex]\Delta[/tex]PE
b) Work done by friction = [tex]\Delta[/tex]KE+[tex]\Delta[/tex]PE
2. 0=[tex]\Delta[/tex]KE+[tex]\Delta[/tex]PE
3. 0=[tex]\Delta[/tex]KE+[tex]\Delta[/tex]PE+[tex]\Delta[/tex]PEspring
1. a) Since no friction, I used 0=[tex]\Delta[/tex]KE+[tex]\Delta[/tex]PE which will become=
0=[1/2(mv^2)f-1/2(m[10^2])]+[mg(hf-hi)]
50m=1/2(mv^2)f+[mg(hf-hi)]
50=1/2(v^2)f+[9.8(0-20)]
100=(v^2)f+[-196]
296=v^2f
v=17.2 m/s
b) Since friction, I used Work done by friction = [tex]\Delta[/tex]KE+[tex]\Delta[/tex]PE
and since I obtained the velocity from part a, I assume I can start off with the velocity I obtained as the initial? (Confused as to whether or not I can do this?)
Work done by friction = [tex]\Delta[/tex]KE+[tex]\Delta[/tex]PE
Ffriction(Cos(180))[tex]\Delta[/tex]x=[1/2(m0^2)f-1/2(m[17.2^2])]+[mg(0-10)]
0=-1/2(m[17.2^2])]+[mg(-10)]-Ffriction(cos(180))25.0
1/2(m[17.2^2])]+[mg(10)]=Ffriction(25.0)
Ffriction=541 Joules
2. Since no friction, I used 0=[tex]\Delta[/tex]KE+[tex]\Delta[/tex]PE again
change in PE will both be the same since the change in height was the same, the bumps does not matter since the difference between the heights are still the same. so this also implies that the change in kinetic energy will also be the same. This then means that the change in KE will be the same between Ramp A and B. However, since Ramp A is pushed with some initial velocity, it starts off with some KE, will be going that much faster because of it.
on second thought, for 3a, these values such as the spring constant and mass wasn't given. They were only given in part 3b. Should I assume I can use them? because if not, i have no idea how to do this problem.
3. a) E1 (energy at spring, only spring potential)=1/2(k)x^2 = 1/2(800)(.03^2) = .36 J
E2 (top of ramp, both KE and PE)=mgh + 1/2(mv^2) = .05(10)(.15)+1/2(.05)v^2
Conservation of energy due to no friction
E1=E2
.36=.075+.025v^2
.285/.025=v^2
v=3.38 m/s @ top of ramp
however, how do i calculate the direction of this?
b) NOT sure
E3 introduced: energy of ball on the floor = mgh + 1/2(mv^2)
E3=E1=E2
.36=.050(10)(-1.0) + 1/2(.050)v^2
.86 = .025v^2
34.4 = v^2
v = 5.87 m/s as it hits the floor?
[negative 1.0 meters for table length because the height of the table that is set to be 0]
some of these i probably did wrong. But i would like a check and some advice is possible. thx.
Thank you and sorry for the long post. I read the book but some concepts are just hard to grasp and visualize
Homework Statement
1. a 55 kg skier skis down a smooth frictionless ramp, slows down over a distance of 25 m on the flat portion at the bottom as shown in the diagram for number 1.
a) The skier pushes off at the top with a initial speed of 10 m/2. Find her speed at the top of the middle hump
b) The skier comes to rest at the end of the rough section. Find the average friction force experienced by the skier
2. You release a frictionless cart at the top of each of the 2 ramps. On ramp B, the cart is released from rest while the Ramp A cart is pushed with some initial velocity. The ramps have the same height as each other. The only difference is that one has a small bump up and the other a bump down. Which cart has the larger speed at the finish? why?
3. a) calculate the velocity of the ball when it just leaves the launching ramp (both magnitude and direction. be sure to specify a coord system.)
b) spring constant k = 800.0 N/m, the spring's compression is 3.00 cm, the mass of the ball is 50.0g, the height of the ramp is 15.0, and H (table's height) is 1.00 meter. With what total speed will the ball hit the floor? (use g = 10 m/s^2)
Homework Equations
1. a) 0=[tex]\Delta[/tex]KE+[tex]\Delta[/tex]PE
b) Work done by friction = [tex]\Delta[/tex]KE+[tex]\Delta[/tex]PE
2. 0=[tex]\Delta[/tex]KE+[tex]\Delta[/tex]PE
3. 0=[tex]\Delta[/tex]KE+[tex]\Delta[/tex]PE+[tex]\Delta[/tex]PEspring
The Attempt at a Solution
1. a) Since no friction, I used 0=[tex]\Delta[/tex]KE+[tex]\Delta[/tex]PE which will become=
0=[1/2(mv^2)f-1/2(m[10^2])]+[mg(hf-hi)]
50m=1/2(mv^2)f+[mg(hf-hi)]
50=1/2(v^2)f+[9.8(0-20)]
100=(v^2)f+[-196]
296=v^2f
v=17.2 m/s
b) Since friction, I used Work done by friction = [tex]\Delta[/tex]KE+[tex]\Delta[/tex]PE
and since I obtained the velocity from part a, I assume I can start off with the velocity I obtained as the initial? (Confused as to whether or not I can do this?)
Work done by friction = [tex]\Delta[/tex]KE+[tex]\Delta[/tex]PE
Ffriction(Cos(180))[tex]\Delta[/tex]x=[1/2(m0^2)f-1/2(m[17.2^2])]+[mg(0-10)]
0=-1/2(m[17.2^2])]+[mg(-10)]-Ffriction(cos(180))25.0
1/2(m[17.2^2])]+[mg(10)]=Ffriction(25.0)
Ffriction=541 Joules
2. Since no friction, I used 0=[tex]\Delta[/tex]KE+[tex]\Delta[/tex]PE again
change in PE will both be the same since the change in height was the same, the bumps does not matter since the difference between the heights are still the same. so this also implies that the change in kinetic energy will also be the same. This then means that the change in KE will be the same between Ramp A and B. However, since Ramp A is pushed with some initial velocity, it starts off with some KE, will be going that much faster because of it.
on second thought, for 3a, these values such as the spring constant and mass wasn't given. They were only given in part 3b. Should I assume I can use them? because if not, i have no idea how to do this problem.
3. a) E1 (energy at spring, only spring potential)=1/2(k)x^2 = 1/2(800)(.03^2) = .36 J
E2 (top of ramp, both KE and PE)=mgh + 1/2(mv^2) = .05(10)(.15)+1/2(.05)v^2
Conservation of energy due to no friction
E1=E2
.36=.075+.025v^2
.285/.025=v^2
v=3.38 m/s @ top of ramp
however, how do i calculate the direction of this?
b) NOT sure
E3 introduced: energy of ball on the floor = mgh + 1/2(mv^2)
E3=E1=E2
.36=.050(10)(-1.0) + 1/2(.050)v^2
.86 = .025v^2
34.4 = v^2
v = 5.87 m/s as it hits the floor?
[negative 1.0 meters for table length because the height of the table that is set to be 0]
some of these i probably did wrong. But i would like a check and some advice is possible. thx.
Thank you and sorry for the long post. I read the book but some concepts are just hard to grasp and visualize
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