A 2-kg block is attached to an ideal spring (K=200 N/m) and initially at rest on

Question

A 2-kg block is attached to an ideal spring (K=200 N/m) and initially at rest on a horizontal frictionless surface. A 100g (.1 kg) ball of clay is thrown at the 2-kg block. The clay is moving horizontally with speed v when it hits and sticks to the block. As a result, the spring compresses a maximum distance of .4 meters.

a) Calculate the energy stored in the spring at maximum compression.

** I used U=1/2(k)(x)^2 and found that U= 16 Joules, but I'm not sure if this is right.

b) Calculate the energy stored in the spring at maximum compression.

** I used 1/2mv^2=1/2kx^2, but I am also not sure if this is correct.

c) Calculate the initial speed v of the clay.

Thank you so much!

Explanation / Answer

a) Energy stored in spring at maximum compression = 1/2 * 200 * 0.4 * 0.4

                                                                                      =    16 J

b) Energy stored in spring at maximum compression = 1/2 * 200 * 0.4 * 0.4

                                                                                      =    16 J

c)    Here, Energy stored in spring = KE lost

     => 16 = 1/2 *   2.1 * v2

      =>   v = 3.9 m/sec     

This is the velocity of block + clay

For initial velocity of clay we have to apply conservation of momentum

=> 0.1 * V = (2 + 0.1) * 3.9

=> V =    81.9 m/sec              ---------------------> initial speed v of the clay.

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