(A) The graph below is of the force exerted by a spring as a function of the dis

Question

(A) The graph below is of the force exerted by a spring as a function of the distance it is stretched from its unstretched length. What is the spring constant of this spring?

1-k = 0.188m/N

2-k = 1.33N/m

3-k = 0.375m/N

4-k = 2.67N/m

5-Not enough information.

(B) The graph below is of the force exerted by a spring as a function of the distance it is stretched from its unstretched length. What is the elastic potential energy if the spring is stretched 1.5m from its equilibrium length?

1-U = 1.5J

2-U = 0.21J

3-No enough information

4-U = 0.42J

5-U = 3.0J

(C) The graph below is of the force exerted by a spring as a function of the distance it is stretched from its unstretched length. Show calculations for each part below. A low-friction cart of mass 5.0kg is attached to the spring, the spring is stretched 1.5m from its equilibrium length, and then the cart is released. What is the kinetic energy of the cart just as the spring passes through its equilibrium length?

1-KE = 1.5J

2-No enough information.

3-KE = 3J

4-KE = 0.42J

5-KE = 0.21J

(D) What is the velocity of the cart at this position?

1-No enough information.

2-v = 0.82m/s

3-v = 0.41m/s

4-v = 2.2m/s

5-v = 1.1m/s

Explanation / Answer

(a)
Force due to Spring, F = k*x
At 3.0 m , Force = 8.0 N
SO,
k = F/x
k = 8/3
k = 2.67 N/m

(b)
Elastic P.E = 1/2 * kx^2

Where k = F/x
k = 2.67

Elastic P.E = 1/2 * 2.67* 1.5^2 J
Elastic P.E = 3.00 J

(C)
At it's Equlibrium Length, All the Elastic P.E gets converted to K.E
Therefore,
Kinetic energy of the cart just as the spring passes through its equilibrium length = 3.00 J

(D)
K.E = 3.0 J
1/2 * m * v^2 = 3.0
v = sqrt(2*3.0/5) m/s
Velocity of Cart, v = 1.1 m/s

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