Question #174793

 A block is pushed against a massless spring, with k = 200 N/m, which is attached to a wall. The spring is compressed 0.04 m before the block is released. As the spring expands back to its original length, what will be the change in kinetic energy of the block?

(A) 8.0 J

(B) 80.0 J

(C) 4.0 J

(D) 0.16 J



Expert's answer

Since the spring is massless, the change in kinetic energy of the block will be equal to the potential energy of the spring in a compressed state according to the energy conservation law. The elastic potential energy is given as follows:


E=kx22E = \dfrac{kx^2}{2}

where k=200N/mk = 200N/m is the spring constant, and x=0.04mx = 0.04m is the compression of the spring. Thus, obtian:


E=2000.0422=0.16JE = \dfrac{200\cdot 0.04^2}{2} = 0.16J

Answer. (D) 0.16 J.


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