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1 year ago

FIND MEFIND V AT THE FIRST HILLFIND HEIGHT OF THE SECOND HILLFIND V AT POINT A

Physics

1 answer:

vitfil [10]1 year ago

7 0

At the starting point, the spring releases potential energy which is converted to kinetic energy of the truck. The formula fr calculating the elastic potential energy of the spring is expressed as

PE = 1/2kx^2

where

x is the extension of the spring

k is the spring constant

From the information given,

k = 8500

x = 7

Thus,

PE = 1/2 x 8500 x 7^2 = 208250 J

Since elastic potential energy of spring = kinetic energy of the truck, it means that

Kinetic energy = 208250

The formula for calculating kinetic energy is expressed as

KE = 1/2mv^2

where

m = mass of truck

v = velocity of truck

From the diagram,

m = 600

Thus,

208250 = 1/2 x 600 x v^2

208250 = 300v^2

v^2 = 208250/300 = 694.17

v = square root of 694.17

v = 26.35 m/s

The velocity at which the truck is moving is 26.35 m/s

The potential energy of the truck at that point is calculated by apply the formula,

Potential energy = mgh

where

g = acceleration due to gravity and its value is 9.81 m/s

h is the height of the truck and it is 20

m is the mass of the truck and it is 600

Thus,

Potential energy = 600 x 9.81 x 20 = 117720

Mechanical energy = potential energy + kinetic energy

Mechanical energy = 208250 + 117720 = 325970 J

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Explanation:

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8 0

2 years ago

Having difficulty finding the PE and KE for these values no mass is given. Does anyone know to go solve these?

Alexandra [31]

11) $1.04\cdot 10^7 J$

12) $1.04\cdot 10^7 J$

13) 50.0 m/s

14) 41.6 m/s

Explanation:

11)

The potential energy of an object is the energy possessed by the object due to its position relative to the ground. It is given by

$PE=mgh$

where

m is the mass of the object

g is the acceleration due to gravity

h is the height relative to the ground

Here in this problem, when the train is at the top, we have:

m = 8325 kg (mass of the train + riders)

$g=9.8 m/s^2$ (acceleration due to gravity)

h = 127 m (height of the train at the top)

Substituting,

$PE=(8325)(9.8)(127)=1.04\cdot 10^7 J$

12)

According to the law of conservation of energy, the total mechanical energy of the train must be conserved (in absence of friction). So we can write:

$KE_t + PE_t = KE_b + PE_b$

where

$KE_t$ is the kinetic energy at the top

$PE_t$ is the potential energy at the top

$KE_b$ is the kinetic energy at the bottom

$PE_b$ is the potential energy at the bottom

The kinetic energy is the energy due to motion; since the train is at rest at the top, we have

$KE_t=0$

Also, at the bottom the height is zero, so the potential energy is zero

$PE_b=0$

Therefore, we find:

$KE_b=PE_t=1.04\cdot 10^7 J$

13)

The kinetic energy of an object is the energy of the object due to its motion. Mathematically, it is given by

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

where

m is the mass of the object

v is the speed of the object

From question 12), we know that the kinetic energy of the train at the bottom is

$KE=1.04\cdot 10^7 J$

We also know that the mass is

m = 8325 kg

Therefore, we can calculate the speed of the train at the bottom:

$v=\sqrt{\frac{2KE}{m}}=\sqrt{\frac{2(1.04\cdot 10^7)}{8325}}=50.0 m/s$

14)

At the top of the second hill, the total mechanical energy of the train is still conserved.

Therefore, we can write again:

$KE_1 + PE_1 = KE_2 + PE_2$

where

$KE_1$ is the kinetic energy at the top of the 1st hill

$PE_1$ is the potential energy at the top of the 1st hill

$KE_2$ is the kinetic energy at the top of the 2nd hill

$PE_2$ is the potential energy at the top of the 2nd hill

From the previous questions, we know that

$KE_1=0$

and

$PE_1=1.04\cdot 10^7 J$

The height of the second hill is

h = 39 m

So we can also find the potential energy at the second hill:

$PE_2=mgh=(8325)(9.8)(39)=3.2\cdot 10^6 J$

So, the kinetic energy at the second hill is

$KE_2=PE_1-PE_2=1.04\cdot 10^7 - 3.2\cdot 10^6 =7.2\cdot 10^6 J$

And so, the speed is

$v=\sqrt{\frac{2KE_2}{m}}=\sqrt{\frac{2(7.2\cdot 10^6)}{8325}}=41.6 m/s$

4 0

3 years ago

He throws a second ball (B2) upward with the same initial velocity at the instant that the first ball is at the ceiling. c. How

Gwar [14]

Answer:

hello your question has some missing parts

A juggler performs in a room whose ceiling is 3 m above the level of his hands. He throws a ball vertically upward so that it just reaches the ceiling.

answer : c) 0.39 sec

d) 2.25 m

e) 1.92 m/sec

Explanation:

The initial velocity of the first ball = 7.67 m/sec ( calculated )

Time required for first ball to reach ceiling = 0.78 secs ( calculated )

Determine how long after the second ball is thrown do the two balls pass each other

Distance travelled by first ball downwards when it meets second ball can be expressed as : d = 1/2 gt^2 = 9.8t^2 / 2

hence d = 4.9t^2 ----- ( 1 )

Initial speed of second ball = first ball initial speed = 7.67 m/sec

3 - d = 7.67t - 4.9t ---- ( 2 )

equating equation 1 and 2

3 = 7.67t therefore t = 0.39 sec

Determine how far the balls are above the Juggler's hands ( when the balls pass each other )

form equation 1 ;

d = 4.9 t^2 = 4.9 *(0.39)^2 = 0.75 m

therefore the height the balls are above the Juggler's hands is

3 - d = 3 - 0.75 = 2.25 m

determine their velocities when the pass each other

velocity = displacement / time

velocity = d / t = 0.75 / 0.39 sec = 1.92 m/sec

7 0

3 years ago

Pls answer 50 points for an answer!!!!

lyudmila [28]

ummmn hi i dont know lol

5 0

2 years ago

At what point does the external energy enter the system?

Phoenix [80]

The correct answer as the first one above !

8 0

3 years ago