In the situation depleted above a 50 g shell ball it placed against a spring. wi
ID: 2060670 • Letter: I
Question
In the situation depleted above a 50 g shell ball it placed against a spring. with a spring; constant of 5000 N/m. which is compressed 5 cm When released the ball will travel a distance of 30 cm to get to the top of the ramp where it will be launched into the air. The ramp is rough over 2/3 of the length the hall traverses, with a frictional coefficient of OS. Once in the air the ball will reach a maximum height and falls down where it lands on a second ramp that is on a platform 1.5 cm above the top of the first ramp. The hall will move along the frictionless platform until it reaches a plunger The plunger will compress until the ball comet to rot The plunger exerts a resistive force. F = 5000x3 N, on the hall. What is the speed of the hall at the top of the first ramp? How high above the first ramp does the hall go? How far is the plunger compressed to stop the hall?Explanation / Answer
(a) m=.05kg, k=5000N/m, xs=.05m, L=0.3m, xf=0.2m, k=0.8, v=?
Eo= -(1/2)(k)(-(xs)2),Ef=(1/2)(m)(v2), Wnc=Ffxf :Eo+|Wnc|=Ef => Eo-Wnc=Ef
(1/2)(k)(xs)2-kmgxf=(1/2)(m)(v2) => [(k)(xs)2-(2)kmgxf]=(m)(v2) => [(k)(xs)2-(2)kmgxf]/[m]=v2=>
v=[[(k)(xs)2-(2)kmgxf]/[m]] => v=[[(5000)(.05)2-(2)(0.8)(.05)(0.2)]/[.05]]=15.8(m/s)
(b) m=.05kg, vo=15.8(m/s),=30o, vx=15.8cos(30), vy=0(m/s), h=?
Eo=(1/2)(m)(vo)2, Ef=(1/2)(m)(vx)2+mgh, Wnc=0J => Eo=Ef
(1/2)(m)(vo)2=(1/2)(m)(vx)2+mgh => (1/2)(m)(vo)2-(1/2)(m)(vx)2=mgh => (1/2)(m)((vo)2-(vx)2)=mgh =>
(1/2)((vo)2-(vx)2)=gh => [(1/2)((vo)2-(vx)2)/(g)]=h => h=[(1/2)((15.8)2-(15.8cos(30))2)/(9.8)]=3.18m
(c) m=.05kg , vo=vx=15.8cos(30)(m/s), h=3.18-1.5=1.68m, vf=0(m/s), k=5000(N/m) xs=?
Eo=(1/2)(m)(vo)2+mgh, Ef=-(1/2)(k)(-(xs)2, Wnc=0J => Eo=Ef
(1/2)(m)(vo)2+mgh=(1/2)(k)(xs)2 => (m)(vo)2+(2)mgh=(k)(xs)2 => [(m)(vo)2+(2)mgh]/[k]=(xs)2 =>
[[(m)(vo)2+(2)mgh]/[k]]=xs, xs=[[(.05)(15.8cos(30))2+(2)(.05)(9.8)(1.68)]/[5000]]=.047m
Now considering that energy was lost due to kinetic firction the compression of the first spring .05m and the compression of the second spring .047m are extremely close only .003m or 3mm difference. This is because friction only subtracted a small amount of energy (.8)(.05)(9.8)(.2)=.0784J. So the answers are reasonable