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A mass, m = 2.1 kg, is ejected horizontally from a compressed coil with a force constant, k = 5.2...

Question:

A mass, {eq}m = 2.1 {/eq} kg, is ejected horizontally from a compressed coil with a force constant, {eq}k = 5.2 {/eq} N/m and compression, {eq}x = 25 {/eq} cm, onto a rough ramp that is {eq}1.5 {/eq} cm long with one end raised to a height, {eq}h= 120 {/eq} cm. The speed at the bottom of the rough ramp is {eq}v = +3.037 {/eq} m/s. What is the average kinetic friction force of the ramp on the mass is (in N)?

Work and Energy:

The mechanical energy of a system is defined as the sum of its kinetic energy of both rotation and translation and its potential energy. When non-conservative forces act on a system, the variation of its mechanical energy is equal to the work of the non-conservative forces acting on it.

Answer and Explanation:

In this case, the only non-conservative force acting on the block is the friction force, therefore:

{eq}\displaystyle\Delta E_m=W_{fnc}\\ \displaystyle\frac{mv^2}{2}-(mgh+\frac{kx^2}{2})=f_kd\cos180^{\circ}\\ \displaystyle f_k=(\frac{mgh}{d}+\frac{kx^2}{2d})-\frac{mv^2}{2d}\\ \displaystyle f_k= \mathrm{(\frac{(2.1\,kg)(10\,m/s^2)(1,20\,m)}{0.15\,m}+\frac{(5.2\,N/m)(0.25\,m)^2}{2(0.15\,m)})-\frac{(2.1\,kg)(3.037\,m/s)^2}{2(0.15\,m)}}\\ \displaystyle f_k= \mathrm{167.6\,N} {/eq}


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Practice Applying Work & Kinetic Energy Formulas

from Physics 101: Help and Review

Chapter 17 / Lesson 4
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