Block Colliding with a Spring Points:3 A movina 3.80 ka block collides with a ho

Question

Block Colliding with a Spring Points:3 A movina 3.80 ka block collides with a horizontal spring whose spring constant is 499 N/m The block compresses the spring a maximum distance of 9.50 cm from its rest position. The coefficient of kinetic friction between the block and the horizontal surface is 0.450. What is the work done by the spring in bringing the block to rest? Consider ??+??-??. what quantities in the problem correspond to each of the variables in the equation? Sabrmt A Incorrect. Tries 1/10 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? Subnit Answer Tries 0/10 What is the speed of the block when it hits the spring? Submit Answer Tries 0/10

Explanation / Answer

Part A -

mass of the block m = 3.8 kg,

spring constant k = 499 N/m

x = 9.50 cm = 0.095 m

spring energy stored in the spring = delta U = (1/2)*k*x^2 = 0.5*499*0.095^2 = 2.25 J

Frictional force, Ff = u*m*g = 0.45*3.8*9.8 = 16.76 N

So, work done by the spring against friction = delta K = Ff * x = 16.76 * 0.095 = 1.60 J

So, work done by the spring to bring the block to rest = 2.25 + 1.60 = 3.85 J

Part B -

delta K = work done against friction = 1.60 J

delta U = spring energy = 2.25 J

delta E = Initial kinetic energy = 3.85 J

Part C -

Mechanical energy dissipated by the force of friction = 1.60 J

Part D -

Total initial energy = kinetic energy of the block = (1/2)*m*v^2

=> (1/2)*m*v^2 = 3.85

=> 0.5*3.80*v^2 = 3.85

=> v^2 = 3.85 / (0.5*3.80)

=> v = 1.42 m/s.

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