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 E = 55.3 J and a maximum displacement from equilibrium of 0.213 m. (a) What is the spring constant? N/m (b) What is the kinetic energy of the system at the equilibrium point? J (c) If the maximum speed of the block is 3.45 m/s, what is its mass? kg (d) What is the speed of the block when its displacement is 0.160 m? m/s (e) Find the kinetic energy of the block at x = 0.160 m. J (f) Find the potential energy stored in the spring when x = 0.160 m. J (g) Suppose the same system is released from rest at x = 0.213 m on a rough surface so that it loses 12.7 J by the time it reaches its first turning point (after passing equilibrium at x = 0). What is its position at that instant?

Explanation / Answer

SIMILAR PROBLEM A horizontal block-spring system with the block on a frictionless surface has total mechanical energy E = 52.6 J and a maximum displacement from equilibrium of 0.204 m. (a) What is the spring constant? N/m 52.6 = 0.5*k*0.204^2 or Force constant k = (2*52.6)/(0.204)^2 = 2.53*10^3 N/m (b) What is the kinetic energy of the system at the equilibrium point? It has to be same as52.6 J because at equlibrium PE is zero and total energy is conserved. (c) If the maximum speed of the block is 3.45 m/s, what is its mass? kg 0.5*m*3.45^2 = 52.6 or m = (2*52.6)/(3.45^2) = 8.8 kg (d) What is the speed of the block when its displacement is 0.160 m? m/s Let it be v. We have 0.5*m*v^2 + 0.5*k*0.160^2 = 52.6 or v^2 = [52.6 - 52.6*(0.160/0.204)^2]/(0.5*8.8) = 4.58 or v = 2.14 m/s (e) Find the kinetic energy of the block at x = 0.160 m. Ke = 0.5*8.8*(2.14^2) J = 20.25 J (f) Find the potential energy stored in the spring when x = 0.160 m. PE = 52.6*(0.160/0.204)^2 J = 32.36 J (g) Suppose the same system is released from rest at x = 0.204 m on a rough surface so that it loses 14.9 J by the time it reaches its first turning point (after passing equilibrium at x = 0). What is its position at that instant? Let it be x. It ism given by 52.6 - 14.9 = 0.5*(2.53*10^3)*x^2 or x^2 = (2*37.7)/(2.53*10^3) = 0.0298 or x = 0.173 m

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