A spring (k = 210 N/m) is fixed at the top of a frictionless plane inclined at a

Q: A spring (k = 210 N/m) is fixed at the top of a frictionless plane inclined at a

A. y = d*sin A = (0.6+0.2)*sin 32 deg = 0.4239 By energy conservation Ki + Ui = Kf + Uf 30 + 0 = Kf + mgy + 0.5*kx^2 = Kf + 1.6*9.81*0.4239 + 0.5*210*0.2^2 30 = kf + 10.8535 kf = 19.1465 J B. Ki + Ui = Kf + Uf Ki + 0 = 0 + mgy1 + 0.5*kx^2 y1 = 1.1*sin 32 deg = 0.5829 Ki = 1.6*9.81*0.5829 + 0.5*210*0.5^2 Ki = 35.39 J comment below if you have any doubt.

ID: 1435371 • Letter: A

Question

A spring (k = 210 N/m) is fixed at the top of a frictionless plane inclined at angle = 32 °. A 1.6 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 30 J. (a) What is the kinetic energy of the block at the instant it has compressed the spring 0.20 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 0.50 m?

Explanation / Answer

y = d*sin A = (0.6+0.2)*sin 32 deg = 0.4239

By energy conservation

Ki + Ui = Kf + Uf

30 + 0 = Kf + mgy + 0.5*kx^2 = Kf + 1.6*9.81*0.4239 + 0.5*210*0.2^2

30 = kf + 10.8535

kf = 19.1465 J

B. Ki + Ui = Kf + Uf

Ki + 0 = 0 + mgy1 + 0.5*kx^2

y1 = 1.1*sin 32 deg = 0.5829

Ki = 1.6*9.81*0.5829 + 0.5*210*0.5^2

Ki = 35.39 J

comment below if you have any doubt.

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A spring (k = 210 N/m) is fixed at the top of a frictionless plane inclined at a

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