angle= 30 degrees A spring is used to propel a block up an inclined plane as sho

Question

angle= 30 degrees

A spring is used to propel a block up an inclined plane as shown in the figure. The horizontal surface along which the block initially moves is frictionless, but ?k = 0.39 for the block on the inclined plane. The block is initially held in place with the spring compressed 100 cm. When released, the block will slide up the ramp and then back down again, eventually recompressing the spring. What is the final compression of the spring (in cm) if the initial compression of the spring was 100 cm? angle= 30 degrees

Explanation / Answer

initially
1/2 * k*x^2= 1/2 *m*vi^2

distance travelled along incline be L cm
so work done by (friction + gravity)=change in KE of mass to zero

(mgcos+mgsin)*L=1/2 *m*vi^2 - 0

Returning:

now the block slips under force of gravity and friction oppose its motion

acceleration down the incline= gsin-gcos

vf^2=u^2+2*a*L

vf^2=0+2*( gsin-gcos)*L

now it compresses spring and its KE conerts to spring energy during compression

1/2*k*xf^2=1/2 *m*vf^2

1/2*k*xf^2=1/2 *m* (2*( gsin-gcos)*L)^2

1/2*k*xf^2=1/2 *m* (2*( gsin-gcos)*(1/2 *m*vi^2 /(mgcos+mgsin)))^2

now put vi=(k/m) * x in above equation

now put values =30 =0.39 g=9.8 x=1 metres

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