Engineering Mechanics: Statics & Dynamics (14th Edition)

Published by Pearson
ISBN 10: 0133915425
ISBN 13: 978-0-13391-542-6

Chapter 14 - Kinetics of a Particle: Work and Energy - Section 14.3 - Principle of Work and Energy for a System of Particles - Problems - Page 195: 1



Work Step by Step

According to the equation of motion $\Sigma F_y=ma_y=0$ $\implies N+Fsin30-mg=0$ $\implies N+100sin30-20(9.81)=0$ $\implies N=146.2N$ Now the work of a constant force is given as $\Sigma U_{1\rightarrow 2}=\Sigma F_x(s_2-s_1)$ $\implies \Sigma U_{1\rightarrow 2}=(Fcos30-\mu_k N)(s_2-s_1)$ $\implies \Sigma U_{1\rightarrow 2}=100cos30-(0.25\times 146.2)(25-15)=500.52N\cdot m$ We know that $\frac{1}{2}mv^2_1+\Sigma U_{1\rightarrow 2}=\frac{1}{2}mv^2_2$ We plug in the known values to obtain: $\frac{1}{2}20\times (8)^2+500.52=\frac{1}{2}\times 20\times v_2^2$ This simplifies to: $v_2=10.67m/s\approx 10.7~m/s$
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