A spring (k = 200 N/m) is fixed at the top of a frictionless plane inclined at a

Question

A spring (k = 200 N/m) is fixed at the top of a frictionless plane inclined at angle ? = 35 (Fig. 1). 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 16 J.

1. What is the kinetic energy of the block at the instant it has compressed the spring 0.20 m?

2. 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?

A spring (k = 200 N/m) is fixed at the top of a frictionless plane inclined at angle ? = 35 (Fig. 1). 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 16 J. 1. What is the kinetic energy of the block at the instant it has compressed the spring 0.20 m? 2. 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

Given K = 200 N/m Theta = 35° m = 1.0 kg d = 0.40 m Ki = 16 J. x = 0.2 m (1) ya = 0.8 * sin(35)=0.458 Ki+vi =Ka+ mgya + 0.5 kx^2 16 = Ka+9.81*0.458+0.5*200*0.2^2 Ka= 16 -4.50-4 = 7.5 J Therefore, a little over 7 J remain as kinetic energy of the block. (2) Ki+vi =Kb+ Vb Vb = mgyb + 0.5 kx^2 = (1.0 m) (9.8 m/s^2) (sin 35) +(1/2) (200 N/m) (0.4m)^2 = 16 J+5.62 J = 21.62 J Thus, the block must be projected with a kinetic energy of 21.62 J if it is to stop when it has compressed the spring by 0.4 m.

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