1. The figure here shows a plot of potential energy U versus position x of a 0.9
ID: 1286957 • Letter: 1
Question
1. The figure here shows a plot of potential energy U versus position x of a 0.90 kg particle that can travel only along an x-axis. (Nonconservative forces are not involved.) Three values are UA = 15 J, UB = 35 J and UC = 45 J. The particle is released at x = 4.5 m with an initial speed of 7.0 m/s, headed in the negative x direction.
(a) If the particle can reach x = 1.0 m, what is its speed there, and if it cannot, what is its turning point?
(b) What are the magnitude and direction of the force on the particle as it begins to move to the left of x = 4.0 m?
Now, suppose, instead, the particle is headed in the positive x direction when it is released at x = 4.5 m at speed 7.0 m/s. (c) If the particle can reach x = 7.0 m, what is its speed there, and if it cannot, what is its turning point? (d) What are the magnitude and direction of the force on the particle as it begins to move to the right of x = 5.0 m?
2. A block of mass m is dropped onto a spring with spring constant k from a height h. The block becomes attached to the spring and compresses it by distance d before momentarily stopping.
(a) Using conservation of mechanical energy, find how far is the spring compressed, i.e. d, when the block comes to rest. Assume gravitational potential energy is zero at y = 0. Your answer must be in terms of m, g, h, and k. Be sure to show all your work.
ANY AND ALL HELP IS GREAT! I'M SO LOST! THANKS IN ADVANCE!!!!! :)
1. The figure here shows a plot of potential energy U versus position x of a 0.90 kg particle that can travel only along an x-axis. (Nonconservative forces are not involved.) Three values are UA = 15 J, UB = 35 J and UC = 45 J. The particle is released at x = 4.5 m with an initial speed of 7.0 m/s, headed in the negative x direction. (a) Using conservation of mechanical energy, find how far is the spring compressed, i.e. d, when the block comes to rest. Assume gravitational potential energy is zero at y = 0. Your answer must be in terms of m, g, h, and k. Be sure to show all your work. ANY AND ALL HELP IS GREAT! I'M SO LOST! THANKS IN ADVANCE!!!!! :) (a) If the particle can reach x = 1.0 m, what is its speed there, and if it cannot, what is its turning point? (b) What are the magnitude and direction of the force on the particle as it begins to move to the left of x = 4.0 m? Now, suppose, instead, the particle is headed in the positive x direction when it is released at x = 4.5 m at speed 7.0 m/s. (c) If the particle can reach x = 7.0 m, what is its speed there, and if it cannot, what is its turning point? (d) What are the magnitude and direction of the force on the particle as it begins to move to the right of x = 5.0 m? 2. A block of mass m is dropped onto a spring with spring constant k from a height h. The block becomes attached to the spring and compresses it by distance d before momentarily stopping.Explanation / Answer
a )
0.5*0.9*(V^2 - 49) = 35-15
V = 9.66
b)
V at x=2
0.5*0.9*(V^2 - 49) = 35-15
V = 9.66
accelration = 11.11 m/s^2
Force in left direction and mag = ma = 10 N
c) 0.5*0.9*(V^2 - 49) = 45-15
V = 10.75 m/s
d )
0.5*0.9*(V^2 - 49 ) = 45-35
V at x=6 is 10.75 m/s
accelration = 57.78m/s^2
Force is righjt side direction and magnitude = 52.003 N
2)
mg(h+d) = 0.5*k*d*d
0.5*k*d^2 - mgd - mgh =0
d = [mg + {(mg)^2 + 2mghk}^0.5 ]/k
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.