3.7 points SerCP11 13.P015. My Notes Ask Your Teacher A horizontal block-spring

Question

3.7 points SerCP11 13.P015. My Notes Ask Your Teacher A horizontal block-spring system with the block on a frictionless surface has total mechanical energy E= 43.2 J and a maximum displacement from equilibrium of 0.199 (a) What is the spring coant? N/m (b) What is the kinetic energy of the system at the equilibrium point? (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 (f) Find the potential energy stored in the spring when x 0.160 m (9) Suppose the same system is released from rest at x0.199 m on a rough surface so that it loses 13.9 J by the time it reaches its first turning point (after passing equilibrium atx 0). What is its position at that instant? Need Help? Read It

Explanation / Answer

(a) Consider the spring constant = k

So, E = 1/2 *k*A^2
k = 2E/A^2
= 2*43.2/0.199^2
= 2182 N/m

(b) At equilibrium, point kinetic energy = total mechanical energy = 43.2 J

(c) Let mass = M
E = (1/2)*M*vmax^2
M = 2*E/vmax^2
= 2 * 43.2/3.45^2
= 7.26 kg

(d) E = (1/2)Mv^2 + (1/2)kx^2
43.2 = (1/2)*7.26*v^2 + (1/2)*2182*0.16^2
43.2 = 3.63 v^2 + 27.9
=> 3.63 v^2 = 15.3
=> v = sqrt(15.3 / 3.63) = 2.05 m/s

(e) KE = (1/2)Mv^2 = (1/2)*7.26*2.05^2 = 15.2 J

(f) PE = E-KE = 43.2 - 15.2 = 28 J

(g) PE1 = (1/2)*k*x1^2 = (1/2) * 2182 * 0.199^2 = 43.20 J
KE1 = 0
KE2 = 0
x2 = ?
Loss of energy = 13.9 J
Therefore KE2 + PE2 = KE1 + PE1 - 13.9
0 + PE2 = 0 + 43.20 - 13.9
PE2 = 29.3 J
(1/2)*k*x2^2 = 29.3
(1/2) * 2182 * x2^2 = 29.3
=> 1091* x2^2 = 29.3
x2 = sqrt(29.3 / 1091) = 0.16 m

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