A spring (75 N/m) has an equilibrium length of 1.00 m. The spring is compressed

Question

A spring (75 N/m) has an equilibrium length of 1.00 m. The spring is compressed to a length of 0.50m and a mass of 2.1kg is placed at its free end on a frictionless slope which makes an angle of 41 degrees with respect to the horizontal. The spring is then released. a) If the mass is not attached to the spring, how far up the slope will the mass move before coming to rest? b) If the mass is attached to the spring, how far up the slope will the mass move before coming to rest? c) Now the incline has a coefficient of kinetic friction. If the block, attached to the spring, is observed to stop just as it reaches the spring's equilibrium position, what is the coefficient of friction?

Explanation / Answer

(a) U=0.5*k*Δx2 When the spring is released its potential energy gets converted into kinetic energy of the mass. When the mass comes to rest all its kinetic energy transforms into potential energy.

∴mgh=0.5*k*Δx2

2.1*9.8*h = 0.5*75*(1-0.50)2

h = 0.45m, in the direction of slope the mass moves 0.45/sin41 = 0.694m

(b) Let the mass moves distance a beyond the eqlb. point of spring in the direction of slope

0.5*k*Δx2 = m*g*(a+0.5)(sin41) + 0.5*k*(a)2

9.375 = 6.75 + 13.5a + 37.5 a2

a= 7/50,-1/2

hence a=0.14m if mass is attached to the spring

Please confirm the two answers, I will post the third in comments if they are right

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