Question

You push a 0.18 kg block at point 1 to point 0, compressing a spring (k = 240 N/m) a distance of 0.20 m. You release the block, and it travels along the track, goes around the first loop (rA= 0.42 m) and then the second loop (rB= 0.89 m)

a)   What is the velocity of the block at point 1 after it is released?

b)   What is the velocity of the block at point A?

c)   What is the velocity of the block at point B?

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Answer 1

Velocity of block after released from point 1 1/2 kx^2 = 1/2 mv^2 240 times 2^2 = 0.18 v^2 v = 7.30 m/s velocity at point �A� By energy conservation - mg(2 r_A) = 1/2 m[v^2_f - v^2_1] - 3.8 times 2 times 0.42 = 1/2 [v^2_f - 7.3^2] v_f = 6.07 m/s velocity at point �B� - 9.0 times m(2 n_B) = 1/2 m[v^2_f - v^2_1] - 9.8 times 2 times 2 times .89 = v^2_f - v^2_1 v^2_f = 18.402 v_f = 4.29 m/s