A spring (k = 190 N/m) is fixed at the top of a frictionless plane inclined at a

Question

A spring (k = 190 N/m) is fixed at the top of a frictionless plane inclined at angle = 39 °. A 1.2 kg block is projected up the plane, from an initial position that is distance d = 0.70 m from the end of the relaxed spring, with an initial kinetic energy of 31 J. (a) What is the kinetic energy of the block at the instant it has compressed the spring 0.30 m? (b) With what kinetic energy must the block be projected up the plane if it is to stop momentarily when it has compressed the spring by 0.40 m?

Explanation / Answer

a) final kinetic energy kf = ki -kinetic energy lost in going up and compressing the spring

kf = 31-1.2*9.8(0.7+0.3)sin39-1/2*190*0.3^2

kf = 31-7.4-8.55 = 15.05 J

b) k = 1.2*9.8(0.7+0.4)sin39+1/2*190*0.4^2 = 8.14+15.2 = 23.34 J

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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