In the figure, a spring with spring constant k = 180 N/m is at the top of a fric

Question

In the figure, a spring with spring constant k = 180 N/m is at the top of a frictionless incline of angle theta = 40 degree. The lower end of the incline is distance D = 0.96 m from the end of the spring, which is at its relaxed length. A 1.9 kg canister is pushed against the spring until the spring is compressed 0.22 m and released from rest, (a) What is the speed of the canister at the instant the spring returns to its relaxed length (which is when the canister loses contact with the spring)? (b) What is the speed of the canister when it reaches the lower end of the incline?

Explanation / Answer

(a)
Using Energy Conservation,
Spring Potential Energy + Initial Potential Energy = Final Potential Energy + Kinetic Energy
1/2*kx^2 +  Initial Potential Energy - Final Potential Energy = 1/2*mv^2
1/2*kx^2 + m*g*x*sin(40) = 1/2*mv^2
Substituing values,
1/2*180*0.22^2 + 1.9*9.8*0.22*sin(40) = 1/2*1.9*v^2
v = 2.71 m/s


(b)
Speed of the cannister when it reaches lower end,
Using Energy Conservation,
Initial Kinetic Energy + Initial Potential Energy = Final Potential Energy + Final Kinetic Energy
1/2*mv^2 + m*g*D*sin(40) = 0 + 1/2*mvf^2
1/2*2.71^2 + 9.8*0.96*sin(40) = 1/2*vf^2
vf =  4.41 m/s

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

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