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 = 41.0 J and a maximum displacement from equilibrium of 0.227 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 = 0.227 m on a rough surface so that it loses 15.7 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

What is the spring constant? N/m
41.0 = 0.5*k*0.227^2 or Force constant k = (2*41.0)/(0.227)^2 = 1.591*10^3 N/m

(b) What is the kinetic energy of the system at the equilibrium point?
It has to be same as 41.0 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 = 41.0 or m = (2*41.0)/(3.45^2) = 6.88 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 = 41.0 or
v^2 = [41.0 - 41.0*(0.160/0.227)^2]/(0.5*6.88) =
v = 2.44 m/s

(e) Find the kinetic energy of the block at x = 0.160 m.
Ke = 0.5*6.88*(2.44^2) J = 20.48 J

(f) Find the potential energy stored in the spring when x = 0.160 m.
PE = 41.0*(0.160/0.227)^2 J = 20.36 J

(g)41.0 = 0.5*k*0.227^2 or Force constant k = (2*41.0)/(0.227)^2 = 1.591*10^3 N/m

(b) It has to be same as 41.0 J because at equlibrium PE is zero and total energy is conserved.

(c)0.5*m*3.45^2 = 41.0 or m = (2*41.0)/(3.45^2) = 6.88 kg

(d) Let it be v. We have 0.5*m*v^2 + 0.5*k*0.160^2 = 41.0 or
v^2 = [41.0 - 41.0*(0.160/0.227)^2]/(0.5*6.88) =
v = 2.44 m/s

(e) Ke = 0.5*6.88*(2.44^2) J = 20.48 J

(f) PE = 41.0*(0.160/0.227)^2 J = 20.36 J

(g) 41.0 - 15.7= 0.5*(1.59*10^3)*x^2
x = 0.0318 m

41.0 - 15.7= 0.5*(1.59*10^3)*x^2
x = 0.0318 m

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