Q.443. Work/Energy Problem

Question 443.
A spring with spring constant k = 195 N/m is fixed at the top of a frictionless plane inclined at an angle θ = 40°. 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 15 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 (from its initial position) if it is to stop momentarily when it has compressed the spring by 0.40 m?

Answer 443.
(a)

Block loses kinetic energy in reaching a height (d + 0.20) sin40° which is equal to its gain of gravitational potential energy
= mg(d + 0.20) sin40° J
= 1 * 9.81 * (0.60 + 0.20) sin40° J
= 5.0446 J
Block also loses kinetic energy in doing work for compressing the spring
= (1/2) kx^2
= (1/2) * 195 * (0.20)^2 J
= 3.9 J
=> its final kinetic energy
= initial kinetic energy - kinetic energy lost in going up and compressing the spring
= 15 - 5.0446 - 3.9 J
= 6.0554 J

(b)
Kinetic energy needed to compress the spring by 0.40 m
= potential energy gained in going up (d + 0.40) sin40° + work done in compressing the spring
= mg(d + 0.40) sin40° + (1/2) k*(0.40)^2 J
= 1 * 9.81 * (0.60 + 0.40) sin40° + (1/2) * 195 * (0.40)^2 J
= 6.31 + 15.6 J
= 21.91 J.

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