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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Work, Power and Energy
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A 2 Kg wood block is on a level board and held against a spring of spring constant k=100 N/m which has been compressed .1 m. The block is released and pushed horizontally across the board. The coefficient of friction between the block and the board is $\mu_k = .2$. Find the velocity of the block just as it leaves the spring.

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Answer: $0.33 m/s$
According to the work-energy theorem, $W = \Delta KE + \Delta PE$
Here, $W = -f_k x = -\mu_k mgx $
$\Delta KE = \large\frac{1}{2}$$ mv_f^2 - 0 = \large\frac{1}{2}$$ m v_f^2$
$\Delta PE = 0 - \large\frac{1}{2}$$kx^2 = -\large\frac{1}{2}$$kx^2$
$\Rightarrow -\mu_k mgx =\large\frac{1}{2}$$ m v_f^2 -\large\frac{1}{2}$$kx^2$
$\Rightarrow -0.2 \times 2 Kg \times 9.8 m/s^2 \times 0.1m = \large\frac{1}{2} $$ (2 Kg \times v_f^2 - 100 N/m \times 0.1^2 m^2)$
Rearranging and solving for $v_f$, $v_f = \sqrt {0.5 - 0.392} m/s = \sqrt{0.108} = 0.33 m/s$
answered Aug 25, 2014 by balaji.thirumalai
 

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