"an ice cube of mass 50.0 g can slide without friction up and down a 25.0 degree slope. the ice cube is pressed against a spring at the bottom of the slope, compressing the spring 0.100 m . the spring constant is 25.0 n/m . when the ice cube is released, how far will it travel up the slope before reversing direction?"

Respuesta :

The elastic potential energy is equal to the gravitational potential energy
 If we set 0 GPE at max spring compression, then
 total E = U = ½kx² = ½ * 25N / m * (0.1m) ² = 0.125 J
 At max height,
 GPE = 0.125 J = mgh = 0.05kg * 9.8m / s² * h
 h = 0.255 m
 Along the slope, that's a distance of
 d = 0.255m / sin25º
 d = 0.604 m
 It's possible they want the distance from the end of the spring - that's 0.504 m

It will travel up the slope as far as about 0.604 m before reversing direction

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Further explanation

Let's recall Elastic Potential Energy formula as follows:

[tex]\boxed{E_p = \frac{1}{2}k x^2}[/tex]

where:

Ep = elastic potential energy ( J )

k = spring constant ( N/m )

x = spring extension ( compression ) ( m )

Let us now tackle the problem!

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Given:

mass of ice cube = m = 50.0 g = 0.05 kg

angle of slope = θ = 25.0°

compression of the spring = x = 0.100 m

spring constant = k = 25.0 N/m

Asked:

distance traveled = d = ?

Solution:

We will use Conservation of Energy formula to solve this problem:

[tex]\texttt{Elastic Potential Energy = Gravitational Potential Energy}[/tex]

[tex]\frac{1}{2}k x^2 = m g h[/tex]

[tex]\frac{1}{2}k x^2 = m g d \sin \theta[/tex]

[tex]d = (\frac{1}{2}k x^2) \div (mg \sin \theta)[/tex]

[tex]d = (\frac{1}{2} \times 25 \times 0.100^2) \div ( 0.05 \times 9.8 \times \sin 25.0^o)[/tex]

[tex]d \approx 0.604 \texttt{ m}[/tex]

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Learn more

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Answer details

Grade: High School

Subject: Physics

Chapter: Elasticity

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