Physics, 16.10.2020 15:01 GreenHerbz206

In a mattress test, you drop a 7.0 kg bowling ball from a height of 1.5 m above a mattress, which as a result compresses 15 cm as the ball comes to a stop. (a) What is the kinetic energy of the ball just bef

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Answer from: Qpaoswp5914

(a) The kinetic energy of the bowling ball just before it hits the matress is 102.974 joules.

(b) The work done by the gravitational force of Earth on bowling ball during the first part of the fall is 102.974 joules.

(d) The work done by the mattress on the bowling ball is 113.272 joules.

Explanation:

The statement is incomplete. The complete question is:

In a mattress test, you drop a 7.0 kg bowling ball from a height of 1.5 m above a mattress, which as a result compresses 15 cm as the ball comes to a stop.

(a) What is the kinetic energy of the ball just before it hits the mattress?

(b) How much work does the gravitational force of the earth do on the ball as it falls, for the first part of the fall (from the moment you drop it to just before it hits the mattress)?

(d) How much work does the mattress do on the ball? (You’ll need to use the results of parts (a) and (c)）

(a) Based on the Principle of Energy Conservation, we know that ball-earth system is conservative, so that kinetic energy is increased at the expense of gravitational potential energy as ball falls:

$K_{1}+U_{g,1} = K_{2}+U_{g,2}$ (Eq. 1)

Where:

$K_{1}$ , $K_{2}$ - Kinetic energies at top and bottom, measured in joules.

$U_{g,1}$ , $U_{g,2}$ - Gravitational potential energies at top and bottom, measured in joules.

Now we expand the expression by definition of gravitational potential energy:

$U_{g,1}-U_{g,2} = K_{2}-K_{1}$

$K_{2}= m\cdot g \cdot (z_{1}-z_{2})+K_{1}$ (Eq. 1b)

Where:

- Mass of the bowling ball, measured in kilograms.

- Gravitational acceleration, measured in meters per square second.

$z_{1}$ , $z_{2}$ - Initial and final heights of the bowling ball, measured in meters.

If we know that $m = 7\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $z_{1}= 1.5\,m$ , $z_{2} = 0\,m$ and $K_{1} = 0\,J$ , the kinetic energy of the ball just before it hits the matress:

$K_{2} = (7\,kg)\cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (1.5\,m-0\,m)+0\,m$

$K_{2} = 102.974\,J$

The kinetic energy of the bowling ball just before it hits the matress is 102.974 joules.

(b) The gravitational work done by the gravitational force of Earth ( $\Delta W$ ), measured in joules, is obtained by Work-Energy Theorem and definition of gravitational potential energy:

$\Delta W = U_{g,1}-U_{g,2}$

$\Delta W = m\cdot g\cdot (z_{1}-z_{2})$ (Eq. 2)

If we know that $m = 7\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $z_{1}= 1.5\,m$ and $z_{2} = 0\,m$ , then the gravitational work done is:

$\Delta W = (7\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (1.5\,m-0\,m)$

$\Delta W = 102.974\,J$

The work done by the gravitational force of Earth on bowling ball during the first part of the fall is 102.974 joules.

(c) The work done by the gravitational force of Earth while the bowling when mattress is compressed is determined by Work-Energy Theorem and definition of gravitational potential energy:

$\Delta W = U_{g,2}-U_{g,3}$

Where $U_{g,3}$ is the gravitational potential energy of the bowling ball when mattress in compressed, measured in joules.

$\Delta W = m\cdot g \cdot (z_{2}-z_{3})$

Where $z_{3}$ is the height of the ball when mattress is compressed, measured in meters.

If we know that $m = 7\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $z_{2}= 0\,m$ and $z_{3} = -0.15\,m$ , the work done is:

$\Delta W = (7\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot [0\,m-(-0.15\,m)]$

$\Delta W = 10.298\,J$

Work done by gravitational force on bowling ball when mattress is compressed is 10.298 joules.

(d) The work done by the mattress on the ball equals the sum of kinetic energy just before mattress compression and the work done by the gravitational force when mattress is compressed:

$\Delta W' = K_{2}+\Delta W$

( $K_{2} = 102.974\,J$ , $\Delta W = 10.298\,W$ )

$\Delta W' = 113.272\,J$

The work done by the mattress on the bowling ball is 113.272 joules.

Answer from: Quest

-first use of geothermal energy occurred more than 10,000 years ago in north america by american paleo-indians

-the first industrial use of geothermal energy began near pisa, italy in late 18th century

-in 1904, italian scientist piero ginori conti invented the first geothermal electric power plant in which steam was used to generate the power

-with the above experiment, the first geothermal plant in usa started in 1922 with a capacity of 250 kilowatts

-however, in 1946 first ground-source geothermal heat pump installed at commonwealth building in portland, oregon

-during the 1960’s, pacific gas and electric began operation of first large scale geothermal power plant in san francisco, producing 11 megawatts

-in 1973, when oil crisis began many countries began looking for renewable energy sources and by 1980’s geothermal heat pumps (ghp) started gaining popularity in order to reduce heating and cooling costs

-as effect of climate change started showing results, governments of various countries joined hands to fight against it, for which kyoto protocol was signed in japan in 1997, laid out emission targets for rich countries and required that they transfer funds and technology to developing countries, 184 countries have ratified it

hope this

Answer from: Quest

Your answer is 32 hope i

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In a mattress test, you drop a 7.0 kg bowling ball from a height of 1.5 m above a mattress, which as...

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