A spring at top of a 45 degreee slope is compressed 0.25 m by a 1.5 kg book. If

Question

A spring at top of a 45 degreee slope is compressed 0.25 m by a 1.5 kg book. If the book is released, it will just reach the edge of the slope at the spring's equilibrium point, and then it will start to slide down the slope. Assume the spring constant is 40 kg/s^2.
a.) If the slope is frictionless, what will be the speed of the book after it has traveled a distance of 10 m in the x direction?
b.) If the book's mass is doubled, does your answer for (a) change, and if so, by how much (b)?

Explanation / Answer

Here conservation of energy takes place i.e., Raise in KE = Loss in PE ==> 0.5 m vo^2 = 0.5 k x^2 ==> vo^2 = (k/m) * x^2 = (40/1.5) * 0.25^2 = 1.6666 m^2/s^2 Therefore v^2 = vo^2 + 2 a s = vo^2 + 2 g sin@ * s = 1.666 + 2 * 9.8 * sin45 * 10 = 140.25 ==> v = 11.84 m/s b) changes. ==> vo^2 = (k/2m) * x^2 = (40/3) * 0.25^2 = 0.8333 m^2/s^2 Therefore v^2 = vo^2 + 2 a s = vo^2 + 2 g sin@ * s = 0.833 + 2 * 9.8 * sin45 * 10 = 139.42 ==> v = 11.80 m/s

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