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nikitadnepr [17]
3 years ago
14

A 35.9 g mass is attached to a horizontal spring with a spring constant of 18.4 N/m and released from rest with an amplitude of

39.5 cm. What is the speed of the mass when it is halfway to the equilibrium position if the surface is frictionless
Physics
2 answers:
lidiya [134]3 years ago
8 0

Answer:

7.74m/s

Explanation:

Mass = 35.9g = 0.0359kg

A = 39.5cm = 0.395m

K = 18.4N/m

At equilibrium position, there's total conservation of energy.

Total energy = kinetic energy + potential energy

Total Energy = K.E + P.E

½KA² = ½mv² + ½kx²

½KA² = ½(mv² + kx²)

KA² = mv² + kx²

Collect like terms

KA² - Kx² = mv²

K(A² - x²) = mv²

V² = k/m (A² - x²)

V = √(K/m (A² - x²) )

note x = ½A

V = √(k/m (A² - (½A)²)

V = √(k/m (A² - A²/4))

Resolve the fraction between A.

V = √(¾. K/m. A² )

V = √(¾ * (18.4/0.0359)*(0.395)²)

V = √(0.75 * 512.53 * 0.156)

V = √(59.966)

V = 7.74m/s

stiks02 [169]3 years ago
7 0

Answer:

The speed of the mass is 7.74 m/s

Explanation:

Given data:

m = 35.9 g = 0.0359 kg

k = 18.4 N/m

A = 39.5 cm = 0.395 m

The displacement in spring is:

x=\frac{A}{2} =\frac{0.395}{2} =0.1975m

The law of conservation of energy:

initial energy = final energy

\frac{1}{2} kA^{2} =\frac{1}{2} kx^{2} +\frac{1}{2} mv^{2} \\kA^{2}=kx^{2}+mv^{2}\\(18.4*0.395^{2} )=(18.4*0.1975^{2} )+(0.0359v^{2} )\\v=7.74m/s

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Read 2 more answers
A physics student hurries through their lab, releasing the bob of a simple pendulum from a height, and allowing it to swing. He
MakcuM [25]

Answer:

The string wasn't taut when he released the bob, causing the bob to move erratically.

The student only timed one cycle, introducing a significant timing error.

The angle of release was too large, so the equation for the period of a pendulum was no longer valid in this case.

Explanation:

The period of a simple pendulum is given by the formula

T=2\pi \sqrt{\frac{L}{g}}

where

T is the period of oscillation

L is the length of the pendulum

g is the acceleration due to gravity

Therefore, it is possible to measure the value of g in an experiment, by taking a pendulum, measuring its length L, and measuring its period of oscillation T. Re-arranging the equation above, we get the value of g as:

g=(\frac{2\pi}{T})^2 L

Here the value of g measured in the experiment is 8 m/s^2 instead of 9.8 m/s^2. Let's now analyze the different options:

The length of the pendulum string was too long, so the equation for the period of a pendulum was no longer valid in this case. --> FALSE. There is no constraint on the length of the pendulum.

The string wasn't taut when he released the bob, causing the bob to move erratically. --> TRUE. This is possible, as if the string is not taut, the pendulum would not start immediately its oscillation, so the period would be larger causing a smaller value measured for g.

The student only timed one cycle, introducing a significant timing error. --> TRUE. This is also impossible: in fact, we can get a more accurate measurement of the period if we measure several oscillations (let's say 10), and then we divide the total time by 10.

The angle of release was too large, so the equation for the period of a pendulum was no longer valid in this case. --> TRUE. The formula written above for the period of the pendulum is valid only for small angles.

The mass of the bob was too large, so the equation for the period of a pendulum was no longer valid in this case. --> FALSE. The equation that gives the period of the pendulum does not depend on the mass.

Instead of releasing the bob from rest, the student threw the bob downward. --> FALSE. In fact, this force would have been applied only at the very first moment, but then later the only force acting on the pendulum is the force of gravity, so the formula of the period would still be valid.

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