A 625-g block is dropped onto a relaxed vertical spring that has a spring consta

Question

A 625-g block is dropped onto a relaxed vertical spring that has a spring constant k =170.0 N/m. The block becomes attached to the spring and compresses the spring 63.7 cm before momentarily stopping. A) While the spring is being compressed, what work is done on the block by the gravitational force on it? B) While the spring is being compressed, what work is done on the block by the spring force? C) What is the speed of the block just before it hits the spring? (Assume that friction is negligible.) D) If the speed at impact is doubled, what is the maximum compression of the spring? I cannot figure out part C and D of this problem

Explanation / Answer

Work done on the block by gravitational force=mgh=0.775*9.8*0.515=3.91 J;......Work done on the spring by spring force=0.5*k*(x^2)=0.5*190*(0.515^2)=25.19 J;........the speed of the block just before it hits the spring be v m/s............ Kinetic energy=0.5*m*v2=25.19-3.91=21.27...So v=7.4 m/s....If the speed is doubled speed=14.8 m/s....So by conserving energies 0.5*m*v2=mgx+0.5*k*(x^2);.....0.5*0.775*14.8*14.8=0.775*9.8*x+0.5*190*(x^2);.....So x=0.9061m=90.61cm

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