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A block with mass m = 7.3 kg is attached to two springs with spring constants k

ID: 1301515 • Letter: A

Question

A block with mass m = 7.3 kg is attached to two springs with spring constants kleft = 30 N/m and kright = 52 N/m. The block is pulled a distance x = 0.25 m to the left of its equilibrium position and released from rest.

1)

What is the magnitude of the net force on the block (the moment it is released)?

N

2)

What is the effective spring constant of the two springs?

N/m

3)

What is the period of oscillation of the block?

s

4)

How long does it take the block to return to equilibrium for the first time?

s

5)

What is the speed of the block as it passes through the equilibrium position?

m/s

6)

What is the magnitude of the acceleration of the block as it passes through equilibrium?

m/s2

7)

Where is the block located, relative to equilibrium, at a time 1.03 s after it is released? (if the block is left of equilibrium give the answer as a negative value; if the block is right of equilibrium give the answer as a positive value)

m

8)

What is the net force on the block at this time 1.03 s? (a negative force is to the left; a positive force is to the right)

N

9)

What is the total energy stored in the system?

J

10)

If the block had been given an initial push, how would the period of oscillation change?

the period would increase

the period would decrease

the period would not change

Explanation / Answer

a)

Force = spring force from left + spring force from right

= 30*0.25 + 52*0.25

= 20.5 N

b)

Effective k = 20.5 / 0.25

= 82 N/m

3)

T = 2*pi*sqrt (m/k)

= 2*3.14*sqrt(7.3 / 82)

= 1.875 s

4)

Time = T/4

= 1.875 / 4

= 0.467 s

5)

At equilbrium, KE = PE_at extreme

1/2*mv^2 = 1/2*ka^2

7.3*v^2 = 82*0.25^2

v = 0.838 m/s

6)

acceleration = 0

7)

w = sqrt (k/m) = sqrt (82 / 7.3) = 3.351 rad/s

x = a*cos (wt)

x = -0.25*cos (3.351*t)

At t = 1.03 s, we get

x = -0.25*cos(3.351*1.03)

= 0.238 m

8)

Spring force from left + spring force from right

= -30*0.238 - 52*0.238

= -19.52 N

9)

Total energy = 1/2*ka^2

= 1/2*82*0.25^2

= 2.5625 J

10)

the period would not change. It only depends upon m and k.

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