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 = 55.0 J and a maximum displacement from equilibrium of 0.268 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 = 0.268 m on a rough surface so that it loses 15.9 J by the time it reaches its first turning point (after passing equilibrium at x = 0). What is its position at that instant? m

Explanation / Answer

part a )

E = 55 J

A = 0.268 m

E = 1/2*kA^2

k = 1531.52 N/m

part b )

At equilibrium, point kinetic energy = total mechanical energy = 55J

part c )

E = 1/2m*vmax^2

m = 2E/vmax^2

m = 9.24 kg

part d )

E = 1/2kx^2 + 1/2mv^2

now displacement = 0.160 m

55 = 19.6J + 1/2*9.24*v^2

v = 2.768 m/s

part e )

KE = 1/2*mv^2

v = 2.768

KE = 35.4 J

part f )

total energy = KE + PE

55 = 35.4 + PE

PE = 19.6 J

part g )

PE initially = 1/2*kx^2 = 55J

PE2 + KE2 = PE1 + KE1 - 15.9

KE1 = 0

KE2 = 0

PE2 = 39.1 J

1/2*kx^2 = 39.1

x = 0.226 m

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