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serg [7]
3 years ago
7

An object of mass 10.0kg is released at point A, slidesto the bottom of the 30° incline, then collides with ahorizontal massless

spring, compressing it a maximumdistance of 0.750m. (See below.) The spring constant is 500N/m, the height of the incline is 2.0 m, and the horizontalsurface is frictionless. (a) What is the speed of the object atthe bottom of the incline? (b) What is the work of frictionon the object while it is on the incline? (c) The springrecoils and sends the object back toward the incline. Whatis the speed of the object when it reaches the base of theincline? (d) What vertical distance does it move back up theincline?
Physics
1 answer:
Elenna [48]3 years ago
6 0

Answer:

Explanation:

Potential energy lost by mass = mgh

= 10 x 9.8 x 2 = 196 J

a ) If v be velocity of mass at the bottom , its kinetic energy will be stored in spring as elastic energy

= 1/2 m v² = 1/2 k x² , k is spring constant , x is compression , m is mass falling down

.5 x 10 v² = .5 x 500 x .75²

v = 5.3 m /s

b ) kinetic energy of mass at the bottom

= /2 m v²

= .5 x 10 x 5.3²

= 140.45 J

energy lost by mass while coming down

=potential energy at the top - kinetic energy at bottom

=  196 - 140.45

= 55.55 J .

This is equal to negative work done by friction

work done by friction = - 55.55 J

c ) Since there will be no loss of energy in compression and extension of spring so , no loss of kinetic energy will take place of mass . So it wil have same velocity that is 5.3 m /s while on its return journey.

d ) kinetic energy at the bottom = 140.45

loss of energy by friction again

= 140.45  - 55.55

= 84.9 J

If h be the height attained

mgh = 84.9

10 x 9.8 x h = 84.9

h = .866 m

( We have assumed that loss of energy in return journey will be same due to friction . )

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Arlecino [84]

(a) The compression of the spring is 0.074 m.

(b) The potential energy of block will be converted to elastic potential energy of the spring when the block comes to rest.

<h3>Compression of the spring [ x ]</h3>

The compression of the spring is calculated as follows;

F = kx

mg = kx

x = mg/k

where;

  • m is mass of the object
  • k is the spring constant

x = (3 x 9.8) / (400)

x = 0.074 m

<h3>Motion of the block</h3>

Based on the principle of conservation of energy, the potential energy of block will be converted to elastic potential energy of the spring when the block comes to rest.

Learn more about elastic potential energy here: brainly.com/question/25996974

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2 years ago
For the wave of light you generated in the Part B, calculate the amount of energy in 1.0 mol of photons with that same frequency
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Answer:

4.79 J

Explanation:

The energy of a single photon is given by

E=hf

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Here we have

f=1.2\cdot 10^{10} Hz

so the energy of one photon is

E_1=(6.63\cdot 10^{-34})(1.2\cdot 10^{10})=7.96\cdot 10^{-24} J

Here we have 1 mol of photons, which contains

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E=NE_1 = (6.022\cdot 10^{23})(7.96\cdot 10^{-24})=4.79 J

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4 years ago
A ball is bouncing to a height that is 1/3 of the previous height after every bounce. What kind of sequence is this? How could y
tamaranim1 [39]

Answer and Explanation:

The ball is bouncing to a height of 1/3 of its previous height this is a type of geometric sequence the total distance can be found by the sum of geometric sequence

For example let the initial height is 243 fit

After one bounce it will reach  243/3 =81 feet

After second bounce 81/3=27 feet

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Here r=\frac{3}{9}=\frac{1}{3}

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