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 = 46.5 J and a maximum displacement from equilibrium of 0.221 m. (a)What is the spring constant? 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? (d)What is the speed of the block when its displacement is 0.160 m? (e)Find the kinetic energy of the block at x=0.160m (f)Find the potential energy stored in the spring when x=0.160 m. (g) Suppose the same system is released from rest at x=0.221 m on a rough surface so that it loses 13.4 J by the time it reaches its first turning point (after passing equiibrium at x=0). What is its position at that instant? (in m.)

Explanation / Answer

Solution: From the question we have:

A horizontal block-spring system with the block on a frictionless surface has total mechanical energy E = 46.5 J and a maximum displacement from equilibrium of 0.221 m.

To find

E=46.5 J
Amplitude A = 0.221 m

A)Let spring constant = k
E = 1/2 *k*A^2
k = 2E/A^2
= 2*46.5/0.221^2
= 1904.13 N/m
Ans: 1904.2 N/m

B) At equilibrium, point kinetic energy = total mechanical energy = 46.5 J
Ans: 46.5 J

C) Let mass = M
E = (1/2)*M*vmax^2
M = 2*E/vmax^2
= 2 * 46.5/3.45^2
= 7.81 kg
Ans: 7.81 kg

D)E = (1/2)Mv^2 + (1/2)kx^2
46.5 = (1/2)*7.81*v^2 + (1/2)*1904.2*0.16^2
46.5 = 3.905 v^2 + 24.37
3.905 v^2 = 22.12
v = sqrt(22.12/3.905) = 2.38 m/s
Ans: 2.38 m/s

E) KE = (1/2)Mv^2 = (1/2)*7.81*2.38^2 = 22.11 J
Ans: 22.11 J

F) PE = E-KE = 46.5 - 22.11 = 24.38 J
Ans: 24.38 J

G) PE1 = (1/2)*k*x1^2 = (1/2) * 1904.2 * 0.221^2 = 46.50 J
KE1 = 0
KE2 = 0
x2 = ?
Loss of energy = 13.4 J
Therefore KE2 + PE2 = KE1 + PE1 - 13.4
0 + PE2 = 0 + 46.50 - 13.4
PE2 = 33.10 J
(1/2)*k*x2^2 = 33.10
(1/2) * 1904.2 * x2^2 = 33.10
952.1 x2^2 = 33.10
x2 = sqrt(33.10/952.1) = 0.18 m
Ans: 0.18 m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.