A spring ( k = 215 N/m) is fixed at the top of a frictionless plane inclined at

Question

A spring (k = 215 N/m) is fixed at the top of a frictionless plane inclined at an angle ? = 40 A spring (k = 215 N/m) is fixed at the top of a frictionless plane inclined at an angle ? = 40 degree (Figure 8-66). A 1.0 kg block is projected up the plane, from an initial position that is distance d = 0.60 m from the end of the relaxed spring, with an initial kinetic energy of 14 J. What is the kinetic energy in Joules of the block at the instant it has compressed the spring 0.20 m? With what kinetic energy in Joules must the block be projected up the plane (from its initial position) if it is to stop momentarily when it has compressed the spring by 0.40 m?

Explanation / Answer

Here

Gain in Potential Energy = (0.60 + 0.20)*9.8*1*Sin(40 degree)

= 5.039 J

Therefore by the Conservation of Energy

Required Kinetic Energy = 14 - 5.039 - 0.5*215*0.20^2

= 4.66 J

Required Kinetic Energy = (0.60 + 0.40)*9.8*1*Sin(40 degree) + 0.5*215*0.40^2

= 23.5 J

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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