The graph below is of the force exerted by a spring as a function of the distanc

Question

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 5.0 kg is attached to the spring, the spring is stretched 1.5 m from its equilibrium length, and then the cart is released. What is the kinetic energy of the cart just as spring passes through its equilibrium length? KE = 3 J KE = 0.42 J KE = 1.5 J KE = 0.21 J No enough information. What is the velocity of the cart at this position ? V = 1.1 m/s v = 0.41 m/s v = 2.2 m/s v= 0.82 m/s No enough information.

Explanation / Answer

Given a graph of Force vs. Position, where you have the force function on the x-axis and the position function on the y-axis, the initial position is 0.0 and then it increases linearly until it reaches 5.0meters where the force in (N) is about 12. So my question is: This graph is for the force exerted by a spring as a function of distance it is stretched from its unstreched length.

a) What is the spring constant for this spring

b) What is the elastic PE if the spring is stretched 1.5 m from its equilibrium length

c) A Low friction cart of mass 5.0kg is attached to the spring, when the spring is stretched 1.5 m 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? What is the velocity of the cart at this position?

• ANS

Given a graph of Force vs. Position, where you have the force function on the x-axis and the position function on the y-axis, the initial position is 0.0 and then it increases linearly until it reaches 5.0meters where the force in (N) is about 12. So my question is: This graph is for the force exerted by a spring as a function of distance it is stretched from its unstreched length.

a) What is the spring constant for this spring

b) What is the elastic PE if the spring is stretched 1.5 m from its equilibrium length

c) A Low friction cart of mass 5.0kg is attached to the spring, when the spring is stretched 1.5 m 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? What is the velocity of the cart at this position?

• ANS

Given a graph of Force vs. Position, where you have the force function on the x-axis and the position function on the y-axis, the initial position is 0.0 and then it increases linearly until it reaches 5.0meters where the force in (N) is about 12. So my question is: This graph is for the force exerted by a spring as a function of distance it is stretched from its unstreched length.

a) What is the spring constant for this spring

b) What is the elastic PE if the spring is stretched 1.5 m from its equilibrium length

c) A Low friction cart of mass 5.0kg is attached to the spring, when the spring is stretched 1.5 m 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? What is the velocity of the cart at this position? (a) By F = - kx
=>k = F/x = 12/5 = 2.4 N/m [ignore the -ve sign as it indicates the direction of the force]
(b) By PE(spring) = 1/2kx^2 = 1/2 x 2.4 x (1.5)^2 = 2.7 J
(c) By PE(spring) = KE(cart) at the spring passes through its equilibrium position
=>KE(cart) = 2.7 J
=>By KE = 1/2mv^2
=>2.7 = 1/2 x 0.5 x v^2
=>v = sqrt[10.8] = 3.29 m/s (a) By F = - kx
=>k = F/x = 12/5 = 2.4 N/m [ignore the -ve sign as it indicates the direction of the force]
(b) By PE(spring) = 1/2kx^2 = 1/2 x 2.4 x (1.5)^2 = 2.7 J
(c) By PE(spring) = KE(cart) at the spring passes through its equilibrium position
=>KE(cart) = 2.7 J
=>By KE = 1/2mv^2
=>2.7 = 1/2 x 0.5 x v^2
=>v = sqrt[10.8] = 3.29 m/s (a) By F = - kx
=>k = F/x = 12/5 = 2.4 N/m [ignore the -ve sign as it indicates the direction of the force]
(b) By PE(spring) = 1/2kx^2 = 1/2 x 2.4 x (1.5)^2 = 2.7 J
(c) By PE(spring) = KE(cart) at the spring passes through its equilibrium position
=>KE(cart) = 2.7 J
=>By KE = 1/2mv^2
=>2.7 = 1/2 x 0.5 x v^2
=>v = sqrt[10.8] = 3.29 m/s

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