Spring Potential Energy
U
s=
1
2
kss
2
Fs=−kss
∆U=−Wint
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Spring Potential Energy - Engineering Physics - Lecture Slides, Slides of Engineering Physics
Here you can find a complete lecture series on Engineering Physics course. These lecture slides includes: Spring Potential Energy, Bound, Real Springs, Horizontal Spring, Negligible Friction, Energy in a Mass Spring System, Path Independence of Potential Energy, Angular Frequency, Rebounding Block
Typology: Slides
2013/2014
1 / 13
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Spring Potential Energy
Us =^1 2
kss^2
Fs = −kss ∆U = −Wint
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Spring Potential Energy Fs = −kss
∆U = −Wint
Us =
2 kss
(^2) − Es docsity.com
Spring Potential Energy
Real Springs
Fs = −kss
Us =
2 kss
(^2) − Es docsity.com
Clicker Quiz
A horizontal spring has a mass attached which can move with negligible friction. You stretch the spring and release the mass from rest. For the resulting motion, which of the fol- lowing statements are TRUE?
- When the spring is (momentarily) fully compressed, K has its largest value.
- When the spring (momentarily) has its relaxed length, U has its largest value.
- When the spring (momentarily) has its relaxed length, K has its smallest value.
- When K is large, U is small, and vice versa.
- When K is large, U is large, and vice versa. docsity.com
Energy in a Mass-Spring System
Us =
2 kss
2 K =^1
2 mv
2
0 2 4 6 8 10 12
U
01
23
45
67
89
10
Energy(J)
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Energy in a Mass-Spring System
Us =
2 kss
2 K =^1
2 mv
2
0 2 4 6 8 10 12
K U
01
23
45
67
89
10
Energy(J)
docsity.com
Clicker Quiz
A ball of mass m is attached to a spring and oscillating with an angular frequency ω. If the motion of the oscillation is given by the following equation, what is the total energy in the oscillator? (^) x(t) = A cos(ωt + φ)
- Etot = 1/ 2 m A^2 ω^2
- Etot = 1/ 2 m A^2 ω^2
- Etot = 1/ 2 m A^2 (ω +φ)^2
- Etot = 1/ 2 m A^2 ω^2 cos^2 (
- Etot = 1/ 2 m A^2 ω^2 sin^2 (
cos(ωt t
- φ)
ωt + φ) ωt + φ)
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Clicker Quiz
A ball of mass m is attached to a spring and oscillating with an angular frequency ω. It has a maximum velocity of vm. What is the magnitude of the force the spring exerts on the mass, when it reaches its maximum amplitude of oscillation?
- Fmax = −k∆x
- Fmax = vmω
- Fmax = mk
- Fmax = mvmω
- Fmax = mvm/ω docsity.com
Path Independence of Potential Energy
A Rebounding Block A metal block of mass 3 kg is moving downward with speed 2 m/s when the bottom of the block is 0.8 mabove the floor (Figure 7.19). When the bottom of the block is 0.4 m above the floor, it strikes the top of a relaxed vertical spring 0.4 m in length. The stiffness of the spring is 2000 N/m.
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