A horizontal block-spring system with the block on a frictionless surface has to

Question

A horizontal block-spring system with the block on a frictionless surface has total mechanical energy E43.3 J and a maximum displacement from equilibrium of 0.207 m. (a) What is the spring constant? 2021.05N/m (b) What is the kinetic energy of the system at the equilibrium point? 4.79 Your response differs from the correct answer by more than 10%. Double check your calculations. J x* (c) If the maximum speed of the block is 3.45 m/s, what is its mass? 7.27578 (d) What is the speed of the block when its displacement is 0.160 m? 4.79 Your response differs from the correct answer by more than 100%, m/s X* (e) Find the kinetic energy of the block at = 0.160 m. 17.41 (f) Find the potential energy stored in the spring when x-0.160 m. 25.88 (g) Suppose the same system is released from rest at x0.207 m on a rough surface so that it loses 13.4 J by the time it reaches its first turning point (after passing equilibrium atx 0). What is its position at that instant? 0.0295 The response you submitted has the wrong sign. m

Explanation / Answer

a) total mechanical energy is E = energy stored in spring

43.3 = 0.5*k*x^2

43.3 = 0.5*k*0.207^2

k = 2021.05 N/m

b) Kinetic energy of teh sytem at the equilibrium point is 43.3 J

c) 0.5*m*Vmax^2 = 43.3

0.5*m*3.45^2 = 43.3

m = 7.28 kg

d) using law of conservation of energy

43.3 = (0.5*m*v^2)+(0.5*k*x^2)

43.3 = (0.5*7.28*v^2)+(0.5*2021.05*0.16^2)

v = 2.18 m/s

g) x = -0.0295 m

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