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Hi! This is already answered. Only have some clarifications! Rate will be given! In a 300-rough incline, a 1.25 kg block is initially at rest beside a spring that is compressed by 24 cm as shown in the figure below. The spring has a spring constant of 180 N/m and the coefficient of sliding friction between the incline and the block is 0.23. When the spring is released, (a) how far up the incline (with respect to its initial position) will the block slide before sliding back? (b) What is the speed of the block at the instant it is 0.34 m from its initial position? Solve by the Work-energy equation.    Note: The 1st image is the drawing for the problem. The 2nd image is the answer.    Just please give explanation regarding the following: a. In Delta Uspring, how do we get the -4.761J. Please write the complete/detailed solutions.

Hi! This is already answered. Only have some clarifications! Rate will be given! In a 300-rough incline, a 1.25 kg block is initially at rest beside a spring that is compressed by 24 cm as shown in the figure below. The spring has a spring constant of 180 N/m and the coefficient of sliding friction between the incline and the block is 0.23. When the spring is released, (a) how far up the incline (with respect to its initial position) will the block slide before sliding back? (b) What is the speed of the block at the instant it is 0.34 m from its initial position? Solve by the Work-energy equation.    Note: The 1st image is the drawing for the problem. The 2nd image is the answer.    Just please give explanation regarding the following: a. In Delta Uspring, how do we get the -4.761J. Please write the complete/detailed solutions.

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Hi! This is already answered. Only have some clarifications! Rate will be given!

In a 300-rough incline, a 1.25 kg block is initially at rest beside a spring that is compressed by 24 cm as shown in the figure below. The spring has a spring constant of 180 N/m and the coefficient of sliding friction between the incline and the block is 0.23. When the spring is released, (a) how far up the incline (with respect to its initial position) will the block slide before sliding back? (b) What is the speed of the block at the instant it is 0.34 m from its initial position? Solve by the Work-energy equation. 

 

Note:

The 1st image is the drawing for the problem.

The 2nd image is the answer. 

 

Just please give explanation regarding the following:

a. In Delta Uspring, how do we get the -4.761J. Please write the complete/detailed solutions.

 

we apply the following equations - WUS + WNE AK+ AU W us work done by force applied by us. W NE work done by non-conservative force A K = change in kinetic energy AU= change in potential energy friction force F₁ = μ₁N= μ₁Mgсose= 0.23x1.25x9.8x √√3/2 F₁=2.4399N r W US=0 WNE - F,xx=-2.4399x AU g=mgh=1.25x9.8xxx sin30=6.125x AK=0 AU =0- spring 2= -4.761J placing values in equation WUS+WNE AK+ AU 0+ (-2.4399x)=0+ (6.125x-4.761) x=0.556m
Transcribed Image Text:we apply the following equations - WUS + WNE AK+ AU W us work done by force applied by us. W NE work done by non-conservative force A K = change in kinetic energy AU= change in potential energy friction force F₁ = μ₁N= μ₁Mgсose= 0.23x1.25x9.8x √√3/2 F₁=2.4399N r W US=0 WNE - F,xx=-2.4399x AU g=mgh=1.25x9.8xxx sin30=6.125x AK=0 AU =0- spring 2= -4.761J placing values in equation WUS+WNE AK+ AU 0+ (-2.4399x)=0+ (6.125x-4.761) x=0.556m
mm 30⁰
Transcribed Image Text:mm 30⁰
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