In Figure (a), a block of mass m lies on a horizontal frictionless surface and i

Question

In Figure (a), a block of mass m lies on a horizontal frictionless surface and is attached to one end of a horizontal spring (spring constant k) whose other end is fixed. The block is initially at rest at the position where the spring is unstretched (x = 0) when a constant horizontal force in the positive direction of the x axis is applied to it. A plot of the resulting kinetic energy of the block versus its position x is shown in Figure (b). The scale of the figure's vertical axis is set by Ks = 3.6 J and the scale of the figure's horizontal axis is set by xmax = 3.2 m.

a) what its the magnitude of vector F?

_______ N

b) What is the value of K?

________ N/m

Explanation / Answer

In Figure (a), a block of mass m lies on a horizontal frictionless surface and is attached to one end of a horizontal spring (spring constant k) whose other end is fixed. The block is initially at rest at the position where the spring is unstretched (x = 0) when a constant horizontal force vector F in the positive direction of the x axis is applied to it. A plot of the resulting kinetic energy of the block versus its position x is shown in Figure (b). The scale of the figure's vertical axis is set by Ks = 2.0 J and the scale of the figure's horizontal axis is set by xmax = 2.4 m.




a) What is the magnitude of F?
b) What is the value of k?

ANS: SIMILAR QUESTION
In Figure (a), a block of mass m lies on a horizontal frictionless surface and is attached to one end of a horizontal spring (spring constant k) whose other end is fixed. The block is initially at rest at the position where the spring is unstretched (x = 0) when a constant horizontal force vector F in the positive direction of the x axis is applied to it. A plot of the resulting kinetic energy of the block versus its position x is shown in Figure (b). The scale of the figure's vertical axis is set by Ks = 2.0 J and the scale of the figure's horizontal axis is set by xmax = 2.4 m.




a) What is the magnitude of F?
b) What is the value of k?

ANS: In Figure (a), a block of mass m lies on a horizontal frictionless surface and is attached to one end of a horizontal spring (spring constant k) whose other end is fixed. The block is initially at rest at the position where the spring is unstretched (x = 0) when a constant horizontal force vector F in the positive direction of the x axis is applied to it. A plot of the resulting kinetic energy of the block versus its position x is shown in Figure (b). The scale of the figure's vertical axis is set by Ks = 2.0 J and the scale of the figure's horizontal axis is set by xmax = 2.4 m.




a) What is the magnitude of F?
b) What is the value of k?

ANS: SIMILAR QUESTION
Use the principle of conservation of energy, the total energy of the block must be the same at the maximum kinetic energy point in the middle as the point of zero kinetic energy at the end.

Total = Kinetic + Potential

Potential Energy of a Spring = ½kx²

½kx² + Kinetic = ½k(x(max))² + (0)

KE = 2 J at x = 1.2 m (half of x(max))

½k(1.2)² + 2 = ½k(2.4)² + (0)

k = 4 / (2.4² - 1.2²)

k = 0.926 N / m <====================

The force exerted by the spring will exactly equal the applied force at maximum deflection...

F = kx = (0.9260(2.4) = 2.22 N <==================
Use the principle of conservation of energy, the total energy of the block must be the same at the maximum kinetic energy point in the middle as the point of zero kinetic energy at the end.

Total = Kinetic + Potential

Potential Energy of a Spring = ½kx²

½kx² + Kinetic = ½k(x(max))² + (0)

KE = 2 J at x = 1.2 m (half of x(max))

½k(1.2)² + 2 = ½k(2.4)² + (0)

k = 4 / (2.4² - 1.2²)

k = 0.926 N / m <====================

The force exerted by the spring will exactly equal the applied force at maximum deflection...

F = kx = (0.9260(2.4) = 2.22 N <================== Use the principle of conservation of energy, the total energy of the block must be the same at the maximum kinetic energy point in the middle as the point of zero kinetic energy at the end.

Total = Kinetic + Potential

Potential Energy of a Spring = ½kx²

½kx² + Kinetic = ½k(x(max))² + (0)

KE = 2 J at x = 1.2 m (half of x(max))

½k(1.2)² + 2 = ½k(2.4)² + (0)

k = 4 / (2.4² - 1.2²)

k = 0.926 N / m <====================

The force exerted by the spring will exactly equal the applied force at maximum deflection...

F = kx = (0.9260(2.4) = 2.22 N <==================

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

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