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alexira [117]
4 years ago
5

A block of mass m is attached to a spring of constant k, which oscillates over a table without friction. Which of the following

assertions is true? Explain why 1. Potential energy of the system is maximum when kinetic energy is also maximum. 2. Potential energy of the system is maximum when block goes through equilibrium position. 3. Kinetic energy of the system is maximum when potential energy is minimum
Physics
1 answer:
Fofino [41]4 years ago
6 0

Answer:

3. Kinetic energy of the system is maximum when potential energy is minimum.

Explanation:

Given that

Mass of block= m

Spring constant =K

Table is friction less.

As we know that in oscillatory motion ,when kinetic energy is maximum then potential energy will become minimum.

At the mean position:

Kinetic energy is maximum.

Potential energy is minimum.

At the extreme position:

Kinetic energy is minimum.

Potential energy is  maximum.

At the mean position velocity of block will be maximum that is why it have maximum kinetic energy and at the extreme position the velocity of block will be minimum that is why it have minimum kinetic energy.

So from above we can say that kinetic energy of the system is maximum when potential energy is minimum.

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Incomplete question as time is missing.I have assumed some times here.The complete question is here

Calculate the displacement and velocity at times of (a) 0.500 s, (b) 1.00 s, (c) 1.50 s, (d) 2.00 s, and (e) 2.50 s for a rock thrown straight down with an initial velocity of 10 m/s from the Verrazano Narrows Bridge in New York City. The roadway of this bridge is 70.0 m above the water.

Explanation:

Given data

Vi=10 m/s

S=70 m

(a) t₁=0.5 s

(b) t₂=1 s

(c) t₃=1.5 s

(d) t₄=2 s

(e) t₅=2.5 s

To find

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Velocity V from t₁ to t₅

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v=v_{i}+gt\\

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S=v_{i}t+(1/2)gt^{2}

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For t₁=0.5 s

v_{1}=v_{i}+gt\\v_{1}=(10m/s)+(9.8m/s^{2} ) (0.5s)\\v_{1}=14.9m/s\\  And\\S_{1} =v_{i}t+(1/2)gt^{2}\\ S_{1}=(10m/s)(0.5s)+(1/2)(9.8m/s^{2} )(0.5s)^{2} \\S_{1}=6.225m

For t₂

v_{2}=v_{i}+gt\\v_{2}=(10m/s)+(9.8m/s^{2} ) (1s)\\v_{2}=19.8m/s\\  And\\S_{2} =v_{i}t+(1/2)gt^{2}\\ S_{2}=(10m/s)(1s)+(1/2)(9.8m/s^{2} )(1s)^{2} \\S_{2}=14.9m

For t₃

v_{3}=v_{i}+gt\\v_{3}=(10m/s)+(9.8m/s^{2} ) (1.5s)\\v_{3}=24.7m/s\\  And\\S_{3} =v_{i}t+(1/2)gt^{2}\\ S_{3}=(10m/s)(1.5s)+(1/2)(9.8m/s^{2} )(1.5s)^{2} \\S_{3}=26.025m

For t₄

v_{4}=v_{i}+gt\\v_{4}=(10m/s)+(9.8m/s^{2} ) (2s)\\v_{4}=29.6m/s\\  And\\S_{4} =v_{i}t+(1/2)gt^{2}\\ S_{4}=(10m/s)(2s)+(1/2)(9.8m/s^{2} )(2s)^{2} \\S_{4}=39.6m

For t₅

v_{5}=v_{i}+gt\\v_{5}=(10m/s)+(9.8m/s^{2} ) (2.5s)\\v_{5}=34.5m/s\\  And\\S_{5} =v_{i}t+(1/2)gt^{2}\\ S_{5}=(10m/s)(2.5s)+(1/2)(9.8m/s^{2} )(2.5s)^{2} \\S_{5}=55.625m

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