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zhuklara [117]
2 years ago
12

A block slides along a frictionless surface with 220 J of kinetic energy. It then goes up a frictionless ramp. When the kinetic

energy drops to 55 J, how much gravitational potential energy does the block have?
Physics
1 answer:
aev [14]2 years ago
8 0

Hello!

For the explanation of this energy conservation exercise, where we'll use <u>energy conservation law</u>, let's see what this principle proposes.

How you should know, mechanical energy conserves in every point, that is to say mechanical energy is same in A point like B point. (Mechanical energy will be represented by "Me")

Once time we know that, let's take the 220 Joules momentum like A point, and when 55 Joules momentum like B point.

Then, let's use the <u>energy conservation principle:</u>

Me(A) = Me(B)

  • We know Mechanical energy in A point, so just lets replace according to our data:

220 J = Me(B)

  • In B point, we know kinetic energy, but <u>we dont know gravitational potential energy</u>, so lets descompose Mechanical energy, into kinetic energy and gravitational potential energy:

220 J = Ke + Gpe

  • We know kinetic energy value, so lets replace it:

220 J = 55 J + Gpe

  • Finally, just clean Gpe and resolve it:

Gpe = 220 J - 55 J = 165 J

Gravitational potential energy is of One hundred sixty five Joules <u>(165 J).</u>

                                                                                  ║Sincerely, ChizuruChan║

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Answer:

A train takes some time to travel 240 km. If the speed had been 20 km per hour more than the one it was carrying, it would have taken 2 hours less to travel this distance. In what time did he cover the 240 km

Explanation:

Given that,

A train travelled a distance of 240km

Let the initial speed be

S_1 = x km/hr

Let assume the time spent on the first journey is

t_1 = a

Now if he increase the speed to

S_2 = (x + 20) km/hr

Then, he would have take 2hrs less time

Then, time t_2 = a - 2

The common data fore the two journey is the distance

Speed = distance / time

For the first stage

S_1 = d / t_1

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x = 240 / a Equation 1

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d = S_2 × t_2

d = (x+20) × (a - 2)

240 = (x+20) × (a - 2). Equation 2

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240 = 240 - 480/a + 20a - 40

240 - 240 + 40 = - 480/a + 20a

240 - 240 + 40 = (-480 + 20a²) / a

40 = (-480 + 20a²) / a

40a = -480 + 20a²

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Divided through by 20

a² - 2a - 24 = 0

a² + 4a - 6a - 24 = 0

a(a+4) -6(a+4) = 0

(a-6)(a+4) = 0

(a-6) = 0 or (a+4) = 0

So, a = 6 or a = -4

The time cannot be negative, then, the time is a = 6hours

So, t_1 = a = 6hours,

So, the time used in the first journey is 6hours

So, in the second journey the time use is 2hours less than the first journey

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Spanish

Un tren recorrió una distancia de 240 km.

Deje que la velocidad inicial sea

S_1 = x km / h

Supongamos que el tiempo dedicado al primer viaje es

t_1 = a

Ahora si aumenta la velocidad a

S_2 = (x + 20) km / h

Entonces, habría tomado 2 horas menos de tiempo

Entonces, el tiempo t_2 = a - 2

Los datos comunes para los dos viajes son la distancia.

Velocidad = distancia / tiempo

Para la primera etapa

S_1 = d / t_1

d = S_1 × a

d = x × a

240 = x • a

x = 240 / a Ecuación 1

Para la etapa dos

d = S_2 × t_2

d = (x + 20) × (a - 2)

240 = (x + 20) × (a - 2). Ecuación 2

Sustituye la ecuación 1 en 2

240 = (240 / a + 20) × (a -2)

240 = 240 - 480 / a + 20a - 40

240 - 240 + 40 = - 480 / a + 20a

240 - 240 + 40 = (-480 + 20a²) / a

40 = (-480 + 20a²) / a

40a = -480 + 20a²

20a² - 40a -480 = 0

Dividido entre 20

a² - 2a - 24 = 0

a² + 4a - 6a - 24 = 0

a (a + 4) -6 (a + 4) = 0

(a-6) (a + 4) = 0

(a-6) = 0 o (a + 4) = 0

Entonces, a = 6 o a = -4

El tiempo no puede ser negativo, entonces, el tiempo es a = 6 horas

Entonces, t_1 = a = 6 horas,

Entonces, el tiempo utilizado en el primer viaje es de 6 horas

Entonces, en el segundo viaje, el uso del tiempo es 2 horas menos que el primer viaje

Entonces, t_2 = 6 - 2 = 4 horas

t_1 = 6 horas

t_2 = 4 horas

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