A moving 1.70 kg block collides with a horizontal spring whose spring constant i

Question

A moving 1.70 kg block collides with a horizontal spring whose spring constant is 264 N/m. A) The block compresses the spring a maximum distance of 8.00 cm from its rest position. The coefficient of kinetic friction between the block and the horizontal surface is 0.300. What is the work done by the spring in bringing the block to rest? B)How much mechanical energy is being dissipated by the force of friction while the block is being brought to rest by the spring? C)What is the speed of the block when it hits the spring?

Explanation / Answer

a) work done by the spring = the potential energy stored in it = 1/2 * k* x^2 = 0.8448 J b) energy dissipated by friction = Mu* N* x Mu=coefficient of friction N=Weight of the block= 1.7x10=17 x= 0.08 meter therefore, energy = 0.408 J c)initial kinetic energy = work done by spring + energy dissipated by friction = 1.2528 J 1/2*m*v^2=1.2528 => v= 1.21 m/sec

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