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3 years ago

8

When the displacement in SHM is equal to 1/4 of the amplitude xm, what fraction of the total energy is (a) kinetic energy and (b

) potential energy? (c) At what displacement, in terms of the amplitude, is the energy of the system half kinetic energy and half potential energy?

1 answer:

Anna71 [15]3 years ago

7 0

Answer:

$\frac{KE}{TE} = \frac{15}{16}$

b)

$\frac{U}{TE} = \frac{1}{16}$

Explanation:

Let the amplitude of SHM is given as A

so the total energy of SHM is given as

$E = \frac{1}{2}kA^2$

now we know that

a)

kinetic energy is given as

$KE = \frac{1}{2}k(A^2 - x^2)$

here

$x = \frac{A}{4}$

so now we have

$KE = \frac{1}{2}k(A^2 - \frac{A^2}{16})$

$KE = \frac{15}{32}kA^2$

now its fraction with respect to total energy is given as

$\frac{KE}{TE} = \frac{15}{16}$

b)

Potential energy is given as

$U = \frac{1}{2}kx^2$

so we have

$U = \frac{1}{32}kA^2$

so fraction of energy is given as

$\frac{U}{TE} = \frac{1}{16}$

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The vertical position at time t is given by

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where

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So substituting it into the equation and solving for t, we find the time of flight of the cargo:

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Answer:

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x = $\frac{m_1 \ 1.2 \ + m \ 0.2 }{M_2}$

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If an object is thrown upward with an initial velocity of 128 ft/second, then its height after t seconds is given by the follow

IrinaK [193]

Answer:

The maximum height attained by the object and the number of seconds are 128 ft and 4 sec.

Explanation:

Given that,

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Now, for maximum height

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$h=128\ ft$

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We know that, when the object reaches ground the height becomes zero

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$t(128-32t)=0$

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Dafna11 [192]

Answers:

a) -2.54 m/s

b) -2351.25 J

Explanation:

This problem can be solved by the <u>Conservation of Momentum principle</u>, which establishes that the initial momentum $p_{o}$ must be equal to the final momentum $p_{f}$ :

$p_{o}=p_{f}$ (1)

Where:

$p_{o}=m_{1} V_{o} + m_{2} U_{o}$ (2)

$p_{f}=(m_{1} + m_{2}) V_{f}$ (3)

$m_{1}=110 kg$ is the mass of the first football player

$V{o}=-7 m/s$ is the velocity of the first football player (to the south)

$m_{2}=75 kg$ is the mass of the second football player

$U_{o}=4 m/s$ is the velocity of the second football player (to the north)

$V_{f}$ is the final velocity of both football players

With this in mind, let's begin with the answers:

a) Velocity of the players just after the tackle

Substituting (2) and (3) in (1):

$m_{1} V_{o} + m_{2} U_{o}=(m_{1} + m_{2}) V_{f}$ (4)

Isolating $V_{f}$ :

$V_{f}=\frac{m_{1} V_{o} + m_{2} U_{o}}{m_{1} + m_{2}}$ (5)

$V_{f}=\frac{(110 kg)(-7 m/s) + (75 kg) (4 m/s)}{110 kg + 75 kg}$ (6)

$V_{f}=-2.54 m/s$ (7) The negative sign indicates the direction of the final velocity, to the south

b) Decrease in kinetic energy of the 110kg player

The change in Kinetic energy $\Delta K$ is defined as:

$\Delta K=\frac{1}{2} m_{1}V_{f}^{2} - \frac{1}{2} m_{1}V_{o}^{2}$ (8)

Simplifying:

$\Delta K=\frac{1}{2} m_{1}(V_{f}^{2} - V_{o}^{2})$ (9)

$\Delta K=\frac{1}{2} 110 kg((-2.5 m/s)^{2} - (-7 m/s)^{2})$ (10)

Finally:

$\Delta K=-2351.25 J$ (10) Where the minus sign indicates the player's kinetic energy has decreased due to the perfectly inelastic collision

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IrinaK [193]

Answer:

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