A block of mass 0.0400 kg is connected to a spring on a horizontal, frictionless

Q: A block of mass 0.0400 kg is connected to a spring on a horizontal, frictionless

given m = 0.0400 kg A = 0.0400 m f = 13.8 Hz a) we know, w = sqrt(k/m) w^2 = k/m ==> k = m*w^2 = m*(2*pi*f)^2 = 0.04*(2*pi*13.8)^2 = 301 N/m b) D c) A d) U_max = (1/2)*k*A^2 = (1/2)*301*0.04^2 = 0.241 J e) KE_max = U_max = 0.241 J f) x = 0 g) 0.241 J

ID: 1870062 • Letter: A

Question

A block of mass 0.0400 kg is connected to a spring on a horizontal, frictionless surface. The block is pulled 0.0400 m to the right of its equilibrium position and released. The frequency of oscillation is observed to be 13.8 Hz (a) What is the spring constant? HINT: Remember that the spring constant is related to the mass and the period (or frequency) N/m SPE KE SPE KE SPE KE SPE KE (b) Which of the energy bar graphs above represents the energy of the block + mass system when the block is passing through the equilibrium position? (c) Which of the energy bar graphs represents the energy of the block + mass system when the block is at a turnaround point?e (d) What is the maximum sping potential energy of the mass + block system? HINT: Use the energy bar graphs to help you.J (e) What is the maximum kinetic energy of the block? HINT: Take a look at the energy bar graphs. (f) What is the position (x) of the block when it has its maximum speed and maximum kinetic energy? HINT: Take another look at those energy bar graphs!m (9) What is the total energy of the block spring system? Remember that the total energy is conserved, so if you know KE SPE at any location, you know the total energy. (h) What is the SPE of the system when the block when it is 0.02 m to the right of its equilibrium position? | (i) What is the KE of the block when it is 0.02 m to the right of its equilibrium position? (Use conservation of energy here!) 0) What is the velocity of the block when it is 0.02 m to the right of its equilibrium position? m/s (K) Why does the answer to (j) include a sign?

Explanation / Answer

given

m = 0.0400 kg

A = 0.0400 m

f = 13.8 Hz

a) we know, w = sqrt(k/m)

w^2 = k/m

==> k = m*w^2

= m*(2*pi*f)^2

= 0.04*(2*pi*13.8)^2

= 301 N/m

b) D

c) A

d) U_max = (1/2)*k*A^2

= (1/2)*301*0.04^2

= 0.241 J

e) KE_max = U_max

= 0.241 J

f) x = 0

g) 0.241 J

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A block of mass 0.0400 kg is connected to a spring on a horizontal, frictionless

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