A 60 kg boy notices a nice fruit on a tree. Even with stretched arms he cannot reach it and is 0.5 m below the fruit. The boy crouches down - his centre of mass descends 0.4m and then he leaps up. Thu

Question

A 60 kg boy notices a nice fruit on a tree. Even with stretched arms he cannot reach it and is 0.5 m below the fruit. The boy crouches down - his centre of mass descends 0.4m and then he leaps up. Thus, he just reaches the fruit. Assume that he exerts a constant force on the ground while he is in contact with the ground. What is this force at least? ( g = 10 m / s − 2 )

TestHub · Accepted Answer

For the last 0.5 m upward motion of man:

$0^{2} = u^{2} - 2 \times g \times 0.5 \Rightarrow u = 10$

For the initial 0.4 m:

$a S v u v^{2} 10 F = \frac{F - 60 g}{60} = 0.4 m = 10 = 0 = u^{2} + 2 a s = 2 (\frac{F}{60} - g) \times 0.4 = 1350 N$

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