A light spring with force constant 3.05 N/m is compressed by 8.48 cm as it is he
ID: 1973197 • Letter: A
Question
A light spring with force constant 3.05 N/m is compressed by 8.48 cm as it is held between a 0.237 kg block on the left and a 0.474 kg block on the right, both resting on a horizontal surface. The spring exerts a force on each block, tending to push them apart. The blocks are simultaneously released from rest. Find the acceleration with which each block starts to move, given that the coefficient of kinetic friction between each block and the surface is the following values. (Let the coordinate system be positive to the right and negative to the left. Be sure to include the sign to indicate the direction of the acceleration.)(a) µ = 0
heavier block m/s2
lighter block m/s2
(b) µ = 0.100
heavier block m/s2
lighter block m/s2
(c) µ = 0.131
heavier block m/s2
lighter block m/s2
Explanation / Answer
Given that Force constant of spring k = 3.05 N/m Compression of the spring x = 8.48 cm Mass of block on left mL = 0.237 kg Mass of block on right mR = 0.474 kg -------------------------------------------------------------------- By Hooke's laws, the force exert by spring Fs = kx = (3.05 N/m) (0.0848 m) = 0.258 N (a) When coefficient of frcition = 0 Then by Newton's laws F = Fs = mLaL Therefore the acceleration of the left block is aL = Fs/mL = -0.258 N / 0.237 kg = -1.08 m/s2 Now for block on right Therefore the acceleration of the right block is aR = Fs/mR = 0.258 N / 0.474 kg = 0.544 m/s2 Now for block on right Therefore the acceleration of the right block is aR = Fs/mR = 0.258 N / 0.474 kg = 0.544 m/s2 aR = Fs/mR = 0.258 N / 0.474 kg = 0.544 m/s2 ------------------------------------------------------------------------------ (b) When coefficient of friction = 0.1 Then by Newton's laws F = Fs -fk = mLaL Therefore the acceleration of the left block is aL = Fs - mLg/mL = -[0.258 N - (0.1)(0.237)(9.8 m/s2) ]/ 0.237 kg = -0.108 m/s2 Now for block on right Therefore the acceleration of the right block is aR = Fs/mR = [0.258 N-(0.1)(0.474)(9.8 m/s2)] / 0.474 kg = -0.435 m/s2 ------------------------------------------------------------------------------------- (c) When coefficient of friction = 0.131 Then by Newton's laws F = Fs -fk = mLaL Therefore the acceleration of the left block is aL = Fs - mLg/mL = -[0.258 N - (0.131)(0.237)(9.8 m/s2) ]/ 0.237 kg = 0.195 m/s2 Now for block on right Therefore the acceleration of the right block is aR = Fs/mR = [0.258 N-(0.131)(0.474)(9.8 m/s2)] / 0.474 kg = -0.739 m/s2 F = Fs -fk = mLaL Therefore the acceleration of the left block is aL = Fs - mLg/mL = -[0.258 N - (0.1)(0.237)(9.8 m/s2) ]/ 0.237 kg = -0.108 m/s2 Now for block on right Therefore the acceleration of the right block is aR = Fs/mR = [0.258 N-(0.1)(0.474)(9.8 m/s2)] / 0.474 kg = -0.435 m/s2 ------------------------------------------------------------------------------------- (c) When coefficient of friction = 0.131 Then by Newton's laws F = Fs -fk = mLaL Therefore the acceleration of the left block is aL = Fs - mLg/mL = -[0.258 N - (0.131)(0.237)(9.8 m/s2) ]/ 0.237 kg = 0.195 m/s2 Now for block on right Therefore the acceleration of the right block is aR = Fs/mR = [0.258 N-(0.131)(0.474)(9.8 m/s2)] / 0.474 kg = -0.739 m/s2 Now for block on right Therefore the acceleration of the right block is aR = Fs/mR = [0.258 N-(0.1)(0.474)(9.8 m/s2)] / 0.474 kg = -0.435 m/s2 ------------------------------------------------------------------------------------- (c) When coefficient of friction = 0.131 Then by Newton's laws F = Fs -fk = mLaL Therefore the acceleration of the left block is aL = Fs - mLg/mL = -[0.258 N - (0.131)(0.237)(9.8 m/s2) ]/ 0.237 kg = 0.195 m/s2 Now for block on right Therefore the acceleration of the right block is aR = Fs/mR = [0.258 N-(0.131)(0.474)(9.8 m/s2)] / 0.474 kg = -0.739 m/s2 ------------------------------------------------------------------------------------- (c) When coefficient of friction = 0.131 Then by Newton's laws F = Fs -fk = mLaL Therefore the acceleration of the left block is aL = Fs - mLg/mL = -[0.258 N - (0.131)(0.237)(9.8 m/s2) ]/ 0.237 kg = 0.195 m/s2 Now for block on right Therefore the acceleration of the right block is aR = Fs/mR = [0.258 N-(0.131)(0.474)(9.8 m/s2)] / 0.474 kg = -0.739 m/s2 Then by Newton's laws F = Fs -fk = mLaL Therefore the acceleration of the left block is aL = Fs - mLg/mL = -[0.258 N - (0.131)(0.237)(9.8 m/s2) ]/ 0.237 kg = 0.195 m/s2 Now for block on right Therefore the acceleration of the right block is aR = Fs/mR = [0.258 N-(0.131)(0.474)(9.8 m/s2)] / 0.474 kg = -0.739 m/s2 F = Fs -fk = mLaL Therefore the acceleration of the left block is aL = Fs - mLg/mL = -[0.258 N - (0.131)(0.237)(9.8 m/s2) ]/ 0.237 kg = 0.195 m/s2 Now for block on right Therefore the acceleration of the right block is aR = Fs/mR = [0.258 N-(0.131)(0.474)(9.8 m/s2)] / 0.474 kg = -0.739 m/s2 Now for block on right Therefore the acceleration of the right block is aR = Fs/mR = [0.258 N-(0.131)(0.474)(9.8 m/s2)] / 0.474 kg = -0.739 m/s2Related Questions
Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.