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Work Energy Theorem

In the given ...

Question

In the given figure, $m_{1} = 200 g,$ $m_{2} = 600 g,$ $μ_{1} = 0.3,$ $μ_{2} = 0.8$ $a n d K = 200 N / m .$ The minimum force that must be applied on $m_{1}$ to just displace $m_{2}$ is (in N). (Take g = 10 m/  $s^{2}$ )

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Solution

Ans: 3 N
$K x_{0} = f_{12}$ , $K x_{0} = μ_{2} m_{2} g$ .
By work energy theorem for $m_{1}$ and spring $F x_{0} - μ_{1} m_{1} g x_{0} - \frac{1}{2} K x_{0}^{2} = 0$ $F = μ_{1} m_{1} g + \frac{K x_{0}}{2} = (μ_{1} m_{1} + \frac{μ_{2} m_{2}}{2}) g = 3 N$

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Similar questions

m_{1} = 200 g,

m_{2} = 600 g,

μ_{1} = 0.3,

μ_{2} = 0.8

a n d K = 200 N / m .

m_{1}

m_{2}

s^{2}

m_{1}

s^{2}

m_{2}

s^{2}

m_{3}

m_{1}

m_{2}

m_{3}

m_{1}

0.6 s

20 kg

40 kg

g = 10 m / s^{2}

m - n = 16

m^{2} + n^{2} = 400

m n

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