Question

(2) A block of mass m = 2kg, slides on a horizontal surface and collides with a spring of force constant 200N/cm. The block compresses the spring 6.16cm from the un-stretched position. Mechanical energy in the amount of 200m) is dissipated due to friction as the block is brought to rest. What was the speed of the block at the instant of collision with the spring? (5-points)

Answer 1

Since non conservative forces are not present,

Therefore we conserve energy before collision and after collision in this case.

Before collision : The block has a kinetic energy due to the velocity it is moving.

After collisio: The block collides with the spring, so all the mechanical energy gets converted to energy due to friction force (200mJ) and spring compression energy.

Before collision

Total KE = 1/2 mv^​2. =1/2(2) v^{​​​​​​2}

After collision,

Energy due to frictional force = 200mJ =0.2 J

Spring energy = 1/2 kx^{​​​​​2} = 1/2(20000) (6.16/100)²

Therefore conserving energy

=> V^{​​​​​​2} = 0.2 + 37.94

=> V = √38.14 = 6.17 m/s