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 = 55.8 J and a maximum displacement from equilibrium of 0.283 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 = 0.283 m on a rough surface so that it loses 12.6 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

E= 0.5*kx^2 55.8 = 0.5*K*0.283^2 K = 1393.5 N/m at equilibrium all energy is converted into KE = 55.8 J E = 0.5*MV^2 55.8 = 0.5*m*3.45^2 m = 9.376 kg E = 0.5*MV^2 + 0.5*kx^2 55.8 = 0.5*9.376*v^2 + 0.5*1393.5*0.16^2 v = 2.85 m/s KE = 0.5*mv^2 = 0.5*9.376*2.85^2 = 38.08 J PE = 55.8 - 38.08 = 17.72 E at that point 55.8 -12.6 = 43.2 J 0.5*kx^2 = 43.2 x^2 = 43.2/(0.5*1393.5) x = 0.249 m

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