1. A spring (k = 75 N/m) has an equilibrium length of 1.00 meter. The spring is compressedto a length of 0.50 m and a mass of 2.0 kg is placed at its free end on a frictionless slopewhich makes an angle of 41° with respect to the horizontal. The spring is then released.0.50 m0 = 41°(a) If the mass is not attached to the spring, how far up the slope will the mass movebefore coming to rest?(b) If the mass is attached to the spring, how far up the slope will the mass move beforecoming to rest?(c) Now the incline has a coefficient of kinetic friction uk. If the block, attached to thespring, is observed to stop as it reaches the spring's equilibrium position, what is µk?

Answered: A spring (k = 75 N/m) has an…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

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1. A spring (k = 75 N/m) has an equilibrium length of 1.00 meter. The spring is compressed
to a length of 0.50 m and a mass of 2.0 kg is placed at its free end on a frictionless slope
which makes an angle of 41° with respect to the horizontal. The spring is then released.
0.50 m
0 = 41°
(a) If the mass is not attached to the spring, how far up the slope will the mass move
before coming to rest?
(b) If the mass is attached to the spring, how far up the slope will the mass move before
coming to rest?
(c) Now the incline has a coefficient of kinetic friction uk. If the block, attached to the
spring, is observed to stop as it reaches the spring's equilibrium position, what is µk?

Expert Solution

Step 1

at highest point the spring energy will be converted into potential energy hence

$\begin{array}{rcl} \frac{1}{2} k {(∆ x)}^{2} & = & m g h \\ h & = & \frac{k}{2 m g} {(∆ x)}^{2} \\ = & \frac{75}{2 \times 2 \times 9.8} {(\frac{1}{2})}^{2} \\ = & 0.4783 m \end{array}$

so the distance up the slope

$d = \frac{0.4783}{\sin (41)} = 0.729 m$

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