A wooden block with mass 1.55 kg is placed against a compressed spring at the bo

Question

A wooden block with mass 1.55 kg is placed against a compressed spring at the bottom of a slope inclined at an angle of 29.0 (point A). When the spring is released, it projects the block up the incline. At point B, a distance of 5.20 m up the incline from A, the block is moving up the incline at a speed of 5.20 m/s and is no longer in contact with the spring. The coefficient of kinetic friction between the block and incline is k = 0.55. The mass of the spring is negligible.

Calculate the amount of potential energy that was initially stored in the spring.

Take free fall acceleration to be 9.80 m/s^2 .

Explanation / Answer

here,

mass of the block , m = 1.55 kg

theta = 29 degree

d = 5.2 m

v = 5.2 m/s

coefficient of friction , uk = 0.55

let the amount of potential energy that was initially stored in the spring be E

E = uk * m * g * cos(theta) * d + 0.5 * m * v^2 + m * g * sin(theta) * d

E = 0.55 * 1.55 * 9.8 * cos(29) * 5.2 + 0.5 * 1.55 * 5.2^2 + 1.55 * 9.8 * sin(29) * 5.2

E = 97.25 J

he amount of potential energy that was initially stored in the spring is 97.25 J

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