problem 1: A) Does a stretched spring have more potential energy than an unstret

Question

problem 1:

A) Does a stretched spring have more potential energy than an unstretched spring?

B) Does a stretched spring have more mass than an unstretched spring? Explain!

C) Can the total amount of energy in a given object ever change? Explain!

please show full steps:

problem 2:

A ball of mass 0.5 kg is placed on a massless platform attached to a vertical spring of spring constant

300 N/m. The spring is compressed a distance of 0.1 m and then released. The ball ?ies o? the spring and reaches

a vertical height H above the uncompressed level of the spring. Find H.

problem 3:

A (sticky) ball of mass 5 kg is dropped from a height of 20 m onto a vertical massless spring of spring

constant k=200 N/m originally in its equilibrium position.

A) What will be the maximum compression of the spring.

B) If the ball sticks to the spring what will be the maximum extension of the spring when the ball rebounds?

Explanation / Answer

A)

Yes it has more energy than unstreched one

B)

No , as streching has nothing to do with change of mass

C)

Yes, it can change, by many methods such as radiation of heat, work done by it,etc

2)

By conservation of energy,

All the P.E of the spring is converted to the gravitational P.E of the ball at height H

So, 0.5*k*x^2 = mg(h) <----here k = spring constant = 300 N/m

, x = compression = 0.1 m

h = height above the compressed spring

So, H = h - 0.1

So, 0.5*300*0.1^2 = 0.5*9.8*h

so, h = 0.306 m

so, H = 0.306-0.1 = 0.206 m <-------------answer

3)

By Similar conservation of energy,

all the gravitaional P.E stored is converted to P.E stored in the spring

So, mgh = 0.5*kx^2 <---here h = height of the object from the compressed position

Note : here h = 20+x <---- 20 m + compression of spring

So, mg(20+x) = 0.5*k*x^2

So , 5*9.8*(20+x) = 0.5*200*x^2

So, x = 3.39 m <-----maximum compression

b)

as the ball rebounds,

using conservation of energy,

all the P.E stored in the spring = potential energy at maximum extension(x') + gravitaional energy gained for extension (x')

So, 0.5*k*x^2 = 0.5*k*x'^2 + mg*(x+x') <---here height gained = x+x'

So, 0.5*200*3.39^2 = 0.5*200*x'^2 + 5*9.8*(3.39+x')

So, x' = 2.9 m <---------------answer

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