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 364 N/m. The block compresses the spring a maximum distance of 5.00 cm from its rest position. The coefficient of kinetic friction between the block and the horizontal surface is 0.340, What is the work done by the spring in bringing the block to rest? Consider AK+AU-AE. What quantities in the problem correspond to each of the variables in the equation? Submit Answer You have entered that answer before Incorrect. Tries 6/12 Previous Tries How much mechanical energy is being dissipated by the force of friction while the block is being brought to rest by the spring? Submit Answer Incorrect. Tries 2/12 Previous Tries What is the speed of the block when it hits the spring? Submit Answer Tries 0/12

Explanation / Answer

work done by spring W = -dU = -(1/2)*k*x^2 =- (1/2)*364*0.05^2 =- 0.455 J <<<<------ANSWER

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dK = change in kinetic energy

dU = change in elastic potential energy of the system

dE = dissipated energy

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mechanical energy dissipated E = uk*m*g*x = 0.34*1.7*9.8*0.05 = 0.283 J

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(K2 - K1) + (U2-U1) = dE

(0-(1/2)*m*v1^2) + ( 0.455 - 0) = -0.283

(0-(1/2)*1.7*v1^2) + ( 0.455 - 0) = -0.283

speed v1 = 0.932 m/s <<<<<------ANSWER

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