Question

A block with mass 0.64 kg is forced against a horizontal spring of negligible mass, compressing the spring a distance of 0.38 m (see figure). The coefficient of kinetic friction between the block and table is 0.410. The Spring constant is k= 100 N/m

a, What is the elastic potential energy stored initially in the spring.

b,How far will block slide along the table before coming to rest?

Answer 1

A)

Hence elastic potential energy stored is 7.22joule

- The given resented as situction can be rep- k=loon/ jenen 2 -* a) Elastic potential energy stored initially in the sprsing is P.E=% kag? PE=yx100x(0.3632 J P.E = 7.22

B)

Hence distance moved by the block after release is 2.8m

y distance moved by the block, x The initial and final kinetic enem are both zero, so that Wooldone by the spring is the negatie of work done by friction, 1 kg? = x mg x x Ykx82 Yumq 7.22 = 2.8 4180 64x9.8 x 22.8m x = 12