First slide

Conservation of mechanical energy

Question

The spring and string shown in the figure are ideal and spring constant of spring is $100 N m^{- 1}$ . The coefficient of friction between 2m and surface is 0.8 and mass m of hanging block is 1kg. The system is released with spring in unstretched position. The maximum extension of spring in cm is __________

Moderate

Solution

Let x= extension in spring then
Distance covered by m is $x_{1} = 2 x$
Loss of P.E of m= gain of P.E of spring + work done by friction
$\begin{array}{l} m_{1} g \times x_{1} = \frac{1}{2} k x^{2} + f \cdot x \\ 1 \times 10 \times 2 x = \frac{1}{2} \times 100 x^{2} + 0.8 \times 2 \times 10 \times x \\ 20 x - 16 x = \frac{1}{2} \times 100 x^{2} \\ 4 x = \frac{1}{2} \times 100 x^{2} \\ x = \frac{8}{100} m = 8 cm \end{array}$

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