explain the solution In a 300-rough incline, a 2.15 kg block is initially at rest beside a spring that is compressed by 42 cm as shown in thefigure below. The spring has a spring constant of 180 N/m and the coefficient of sliding friction between the incline andthe block is 0.32. When the spring is released, (a) how far up the incline (with respect to its initial position) will the blockslide before sliding back? (b) What is the speed of the block at the instant it is 0.43 m from its initial position? Solve bythe Work-energy equation.1ww30⁰Transcribed Image Text: In a 300-rough incline, a 2.15 kg block is initially at rest beside a spring that is compressed by 42 cm as shown in thefigure below. The spring has a spring constant of 180 N/m and the coefficient of sliding friction between the incline and the block is 0.32. Whenthe 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 speedof the block at the instant it is 0.43 m from its initial position? Solve by the Work-energy equation. ww 30° Free-Body DiagramFsmg cosov30°ySin 30= 4भmg sinomg30°Mass, m= 2.15 kgAngle Q=30°{ y = x/2Compression, 5=0.42mSpring Constant, K= 180 N/mfriction coefficient 4x= 0·32Calculating spring force,fs KXS =Calculating the friction force,f₁== xx my coso-0.3242-15x9-8x (08302 8.26 NThe energy loss by the block to slidхEloss = fix=8-26xAt the top the potential energy is.P.Etop= mghApplying conservation of energy:115² = mgly + Eloss8.26x180x (0-42)² = 215x9.8 xy +:: Y= x/2x=84.5 cm(6)Answerwhen x=0.43m0.43Applying conservation of energy1 mv² = 1 ks²_mgy - Fixy215²x100x (0-42) ² - 1.25 x 9.8 x 0.43 8.26x 0-43Vz 204 mls2= 180x 0.42=75-6N

Name: explain the solution In a 300-rough incline, a 2.15 kg block is initia
Uploaded: 2024-03-20T06:57:25.000Z
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Description: explain the solution In a 300-rough incline, a 2.15 kg block is initially at rest beside a spring that is compressed by 42 cm as shown in thefigure below. The spring has a spring constant of 180 N/m and the coefficient of sliding friction between the