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VashaNatasha [74]
3 years ago
8

When the displacement of a mass on a spring is 1/2a the half of the amplitude, what fraction of the mechanical energy is kinetic

energy?
Physics
2 answers:
Serhud [2]3 years ago
6 0

Answer:

KE : TE = 3 : 4

Explanation:

As we know that the total mechanical energy of the object which is executing SHM is given by

E_{total} = \frac{1}{2}KA^2

here we know that

A = amplitude of SHM

K = spring constant

now we know that total mechanical energy of the spring is always constant so here we can say

kinetic energy + Potential energy = total mechanical energy

we know that potential energy of the spring at any given distance is

U = \frac{1}{2}kx^2

at given position x = A/2

U = \frac{1}{2}K(\frac{A}{2})^2 = \frac{1}{8}KA^2

now we have

KE + \frac{1}{8}KA^2 = \frac{1}{2}KA^2

KE = \frac{3}{8}KA^2

now ratio of kinetic energy and total mechanical energy will be given as

KE : TE = \frac{3}{8}KA^2 : \frac{1}{2}KA^2

KE : TE = 3 : 4

USPshnik [31]3 years ago
5 0
Total energy is a spring:
E = \frac{1}{2} kx^2 +  \frac{1}{2} mv^2 =  \frac{1}{2} ka^2

At x = 0.5a:
\frac{1}{2} k \frac{a}{2} ^2 +  \frac{1}{2} mv^2 =  \frac{1}{2} ka^2 \\  \frac{1}{2} mv^2 =  \frac{1}{2} ka^2 -  \frac{1}{8} ka^2 =  \frac{3}{8} ka^2

The ration:
\frac{ \frac{3}{8}ka^2 }{ \frac{1}{2} ka^2} =  \frac{3}{4}
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two engines are turned on for 763 s at a moment when the velocity of the craft has x and y components of v0x = 6380 m/s and v0y
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