Physics: What is the potential energy of a 25 kg rock that is on the ground

Potential energy is the energy stored in an object by the virtue of its position. Potential energy...

What is the potential energy of a 25 kg rock, №15224909, 30.10.2020 11:52

Physics

During a rockslide, a 670 kg rock slides from rest down a hillside that is 740 m along the slope and 240 m high. The coefficient of kinetic friction between the rock and the hill surface is 0.25. (a) If the gravitational potential energy U of the rock-Earth system is zero at the bottom of the hill, what is the value of U just before the slide? (b) How much energy is transferred to thermal energy during the slide? (c) What is the kinetic energy of the rock as it reaches the bottom of the hill? (d) What is its speed then?

Step-by-step answer

P Answered by PhD

a) $1.58\cdot 10^6 J$

b) $1.15\cdot 10^6 J$

c) $0.43\cdot 10^6 J$

d) 35.8 m/s

Explanation:

a)

The gravitational potential energy of an object is the energy possessed by the object due to its location with respect to the ground.

It is given by:

U=mgh

where

m is the mass of the object

g is the acceleration due to gravity

h is the height of the object, relative to a reference level

Here, the reference level is taken at the bottom of the hill (where the potential energy is zero).

So, we have:

m = 670 kg is the mass of the rock

g=9.8 m/s^2

h = 240 m is the initial height of the rock

So, the potential energy of the rock just before the slide is

$U=(670)(9.8)(240)=1.58\cdot 10^6 J$

b)

The energy transferred to thermal energy during the slide is equal to the work done by friction, which is:

W=F_f d

where

F_f is the force of friction

d = 740 m is the displacement of the rock along the ramp

The force of friction is given by:

$F_f=-\mu mg cos \theta$

where

$\mu=0.25$ is the coefficient of friction

m = 670 kg is the mass of the rock

$\theta$ is the angle of the ramp

Since we know the lenght of the ramp (d = 740 m) and the height (h = 240 m), we can find the angle:

$\theta=sin^{-1}(\frac{h}{d})=sin^{-1}(\frac{240}{740})=18.9^{\circ}$

Therefore, the work done by friction is:

$W=-\mu m g cos \theta d =-(0.25)(670)(9.8)(cos 18.9^{\circ})(740)=-1.15\cdot 10^6 J$

So, the energy transferred to thermal energy is $1.15\cdot 10^6 J$ .

c)

According to the law of conservation of energy, the kinetic energy of the rock as it reaches the bottom of the hill will be equal to the initial potential energy (at the top) minus the energy transformed into thermal energy.

Therefore, we have:

K_f = U_i -E_t

where here we have:

$U_i=1.58\cdot 10^6 J$ is the potential energy of the rock at the top of the hill

$E_t=1.15\cdot 10^6 J$ is the energy converted into thermal energy

Substituting, we find

$K_f=1.58\cdot 10^6-1.15\cdot 10^6=0.43\cdot 10^6 J$

So, this is the kinetic energy of the rock at the bottom of the hill.

d)

The kinetic energy of the rock at the bottom of the hill can be rewritten as

$K_f=\frac{1}{2}mv^2$

where

m is the mass of the rock

v is its final speed

In this problem, we have:

$K_f=0.43\cdot 10^6 J$ is the final kinetic energy of the hill

m = 670 kg is the mass of the rock

Therefore, the final speed of the rock is:

$v=\sqrt{\frac{2K_f}{m}}=\sqrt{\frac{2(0.43\cdot 10^6)}{670}}=35.8 m/s$

Physics

Arock at the top of a 30 meter cliff has a mass of 25 kg. calculate the rock’s gravitational potential energy when dropped off the cliff. assuming that energy is conserved and there is no air friction and gravity is 9.8 m/sec., at the bottom of the cliff the rock’s potential energy is completely converted to kinetic energy. use the formula for kinetic energy to calculate the rock’s speed at the bottom of the cliff. show all calculations.

Step-by-step answer

P Answered by PhD

4

Lets see:-

We have our formula for potential energy, which we are trying to solve for,

PE = mgh or potential energy = mass * gravity * height

So we know that PE all depends on these.

Height : 30 meters
Mass : 25 kg
Gravity (which is always constant) : 9.8 m/s/s

Now add into formula.

PE = 25*30*9.8
PE = 7350 Joules

PE = 7350 Joules

Physics

Arock at the top of a 30 meter cliff has a mass of 25 kg. calculate the rock’s gravitational potential energy when dropped off the cliff. assuming that energy is conserved and there is no air friction and gravity is 9.8 m/sec., at the bottom of the cliff the rock’s potential energy is completely converted to kinetic energy. use the formula for kinetic energy to calculate the rock’s speed at the bottom of the cliff. show all calculations.

Step-by-step answer

P Answered by PhD

4

Lets see:-

We have our formula for potential energy, which we are trying to solve for,

PE = mgh or potential energy = mass * gravity * height

So we know that PE all depends on these.

Height : 30 meters
Mass : 25 kg
Gravity (which is always constant) : 9.8 m/s/s

Now add into formula.

PE = 25*30*9.8
PE = 7350 Joules

PE = 7350 Joules

Physics

During a rockslide, a 670 kg rock slides from rest down a hillside that is 740 m along the slope and 240 m high. The coefficient of kinetic friction between the rock and the hill surface is 0.25. (a) If the gravitational potential energy U of the rock-Earth system is zero at the bottom of the hill, what is the value of U just before the slide? (b) How much energy is transferred to thermal energy during the slide? (c) What is the kinetic energy of the rock as it reaches the bottom of the hill? (d) What is its speed then?

Step-by-step answer

P Answered by PhD

a) $1.58\cdot 10^6 J$

b) $1.15\cdot 10^6 J$

c) $0.43\cdot 10^6 J$

d) 35.8 m/s

Explanation:

a)

The gravitational potential energy of an object is the energy possessed by the object due to its location with respect to the ground.

It is given by:

U=mgh

where

m is the mass of the object

g is the acceleration due to gravity

h is the height of the object, relative to a reference level

Here, the reference level is taken at the bottom of the hill (where the potential energy is zero).

So, we have:

m = 670 kg is the mass of the rock

g=9.8 m/s^2

h = 240 m is the initial height of the rock

So, the potential energy of the rock just before the slide is

$U=(670)(9.8)(240)=1.58\cdot 10^6 J$

b)

The energy transferred to thermal energy during the slide is equal to the work done by friction, which is:

W=F_f d

where

F_f is the force of friction

d = 740 m is the displacement of the rock along the ramp

The force of friction is given by:

$F_f=-\mu mg cos \theta$

where

$\mu=0.25$ is the coefficient of friction

m = 670 kg is the mass of the rock

$\theta$ is the angle of the ramp

Since we know the lenght of the ramp (d = 740 m) and the height (h = 240 m), we can find the angle:

$\theta=sin^{-1}(\frac{h}{d})=sin^{-1}(\frac{240}{740})=18.9^{\circ}$

Therefore, the work done by friction is:

$W=-\mu m g cos \theta d =-(0.25)(670)(9.8)(cos 18.9^{\circ})(740)=-1.15\cdot 10^6 J$

So, the energy transferred to thermal energy is $1.15\cdot 10^6 J$ .

c)

According to the law of conservation of energy, the kinetic energy of the rock as it reaches the bottom of the hill will be equal to the initial potential energy (at the top) minus the energy transformed into thermal energy.

Therefore, we have:

K_f = U_i -E_t

where here we have:

$U_i=1.58\cdot 10^6 J$ is the potential energy of the rock at the top of the hill

$E_t=1.15\cdot 10^6 J$ is the energy converted into thermal energy

Substituting, we find

$K_f=1.58\cdot 10^6-1.15\cdot 10^6=0.43\cdot 10^6 J$

So, this is the kinetic energy of the rock at the bottom of the hill.

d)

The kinetic energy of the rock at the bottom of the hill can be rewritten as

$K_f=\frac{1}{2}mv^2$

where

m is the mass of the rock

v is its final speed

In this problem, we have:

$K_f=0.43\cdot 10^6 J$ is the final kinetic energy of the hill

m = 670 kg is the mass of the rock

Therefore, the final speed of the rock is:

$v=\sqrt{\frac{2K_f}{m}}=\sqrt{\frac{2(0.43\cdot 10^6)}{670}}=35.8 m/s$

Physics

1. calculate the kinetic energy of a car with a mass of 500 kg traveling at a velocity of 10 m/s north. a. 2,500 j b. 50,000 j c. 100 j d. 25,000 j 2. a rock has a mass of 25 kg. the acceleration due to gravity is 10 m/s^2 and the height of the height of the rock is 42 m. find the potential energy of the rock. a. 1,050 j b. 5,250 j c. 220,500 j d. 10,500 j 3. thermal equilibrium is reached when a. objects have enough kinetic energy for them to be at terminal velocity b. the sound vibrations reach a high pitch c. two objects in contract with each

Step-by-step answer

P Answered by Specialist

6

1) Option D

2) Option D

3) Option C

Explanation:

1) Kinetic energy $=\frac{1}{2} mv^2$

m = 500 kg

v = 10 m/s

Kinetic energy $=\frac{1}{2} *500*10^2=25000 J$

Option D is the correct answer.

2) Potential energy = mgh

m = 25 kg

g = 10 m/s²

h = 42 m

Potential energy = 25 * 10 * 42 = 10500 J

Option D is the correct answer.

3) Thermal equilibrium means no heat flow from each other two systems, that is achieved when both systems have same temperature.

Option C is the correct answer.

Physics

1. calculate the kinetic energy of a car with a mass of 500 kg traveling at a velocity of 10 m/s north. a. 2,500 j b. 50,000 j c. 100 j d. 25,000 j 2. a rock has a mass of 25 kg. the acceleration due to gravity is 10 m/s^2 and the height of the height of the rock is 42 m. find the potential energy of the rock. a. 1,050 j b. 5,250 j c. 220,500 j d. 10,500 j 3. thermal equilibrium is reached when a. objects have enough kinetic energy for them to be at terminal velocity b. the sound vibrations reach a high pitch c. two objects in contract with each

Step-by-step answer

P Answered by Master

6

1) Option D

2) Option D

3) Option C

Explanation:

1) Kinetic energy $=\frac{1}{2} mv^2$

m = 500 kg

v = 10 m/s

Kinetic energy $=\frac{1}{2} *500*10^2=25000 J$

Option D is the correct answer.

2) Potential energy = mgh

m = 25 kg

g = 10 m/s²

h = 42 m

Potential energy = 25 * 10 * 42 = 10500 J

Option D is the correct answer.

3) Thermal equilibrium means no heat flow from each other two systems, that is achieved when both systems have same temperature.

Option C is the correct answer.

Physics

Idon t understand these following questions, can someone me? 1) find the work done by a 25 n force applied for 6 meters 2) calculate the potential energy of a 5 kg object sitting on a 3 m ledge 3) a rock is at the top of a 20 meter tall hill, the rock has a mass of 10 kg. how much potential energy does it have?

Step-by-step answer

P Answered by PhD

10

These questions are testing to see if you know
the formulas for work and potential energy ... or
if you know where to find them when you need them.

1). The formula for work is:

   Work = (force) x (distance) .

In this question: Work = (25 N) x (6 m) = 160 joules .

The formula for potential energy is

   PE = (mass) x (gravity) x (height) .

On Earth, gravity = 9.8 m/s²

This information gives you both of the other questions.

#2). PE = (mass) x (gravity) x (height)

       = (5 kg) x (9.8 m/s²) x (3m) = 147 joules

#3). PE = (mass) x (gravity) x (height)

       = (10 kg) x (9.8 m/s²) x (20 m) = 1,960 joules

Physics

1. Determine the potential energy of a 2kg rock at the top of a hill that is 20m high. 2. Determine the kinetic energy of a 2000 kg roller coaster car that is moving with a speed of 25 m/s. 3. A 80kg freezer is located in an office on the 59 floor of an office building 200 meters above the ground. What is the potential energy of the freezer?

Step-by-step answer

P Answered by Specialist

$1)\:392\:\text{J}\:(400\:\text{J with one significant figure)},\\2)\:625,000\:\text{J}\:(600,000\:\text{J with one significant figure)},\\3)\:156,800\:\text{J}\:(200,000\:\text{J with one significant figure)},$

Explanation:

1. The potential energy of an object is given by PE=mgh . Substituting given values, we have:

$PE=2\cdot 9.8\cdot 20=\boxed{392\:\text{J}}$

2. The kinetic energy of an object is given by $KE=\frac{1}{2}mv^2$ . Substituting given values, we have:

$KE=\frac{1}{2}\cdot 2000\cdot 25^2 =\boxed{625,000\: \text{J}}$

3. 1. The potential energy of an object is given by PE=mgh . Substituting given values, we have:

$PE=80\cdot 9.8\cdot 200=\boxed{156,800\:\text{J}}$

Physics

Idon t understand these following questions, can someone me? 1) find the work done by a 25 n force applied for 6 meters 2) calculate the potential energy of a 5 kg object sitting on a 3 m ledge 3) a rock is at the top of a 20 meter tall hill, the rock has a mass of 10 kg. how much potential energy does it have?

Step-by-step answer

P Answered by PhD

10

These questions are testing to see if you know
the formulas for work and potential energy ... or
if you know where to find them when you need them.

1). The formula for work is:

   Work = (force) x (distance) .

In this question: Work = (25 N) x (6 m) = 160 joules .

The formula for potential energy is

   PE = (mass) x (gravity) x (height) .

On Earth, gravity = 9.8 m/s²

This information gives you both of the other questions.

#2). PE = (mass) x (gravity) x (height)

       = (5 kg) x (9.8 m/s²) x (3m) = 147 joules

#3). PE = (mass) x (gravity) x (height)

       = (10 kg) x (9.8 m/s²) x (20 m) = 1,960 joules

Physics

1. a rock is falling off a cliff. the work that gravity performs on the rock is a) zero b)positive c)negative 2. the sunlight that reaches earth is generated by a transformation of a) mechanical energy b) atmospheric energy c)chemical energy d)nuclear energy 3.at a department store, a toy train travels on a loop of track of length 14 m. the coefficient of rolling friction between the train s wheels and the track is 0.10, and the weight of the train is 65 n. how much work does the train do against friction during each trip around the loop? a) 91

Step-by-step answer

P Answered by PhD

4

1). A rock is falling off the cliff so here in this case force of gravity will act downwards as well as it will displace downwards so here both are along same direction

When force and displacement both are along same direction so work done is always positive.

so here answer will be (B) POSITIVE

2). In case of sunlight the energy is produced on the surface of sun which is radiated in all directions

This energy is the result of nuclear reactions that occurs on the surface of sun

So answer will be (D) NUCLEAR ENERGY

3). Work done by friction is given as

$W = F_f \times distance$