%3Cdiv%20class%3D%22c2%22%3E%0A%3Cdiv%20id%3D%22MathJax_MSIE_Frame%22%20class%3D
ID: 2297556 • Letter: #
Question
%3Cdiv%20class%3D%22c2%22%3E%0A%3Cdiv%20id%3D%22MathJax_MSIE_Frame%22%20class%3D%22c1%22%3E%3C%2Fdiv%3E%0A%3C%2Fdiv%3E%0A%3Cp%3E1)%20A%20man%202kg%20drops%20a%20brick%20from%20the%20top%20of%20a%20100m%20tall%20building.%0Ahow%20fast%20does%20it%20hit%20the%20ground%3F%3C%2Fp%3E%0A%3Cp%3E%20%3C%2Fp%3E%0A%3Cp%3E2)A%20box%20of%20mass%2010kg%20on%20an%20inclined%20plane%20at%2030degree%2C%20with%20a%0Akinetic%20coefficient%20friction%20of%203.0%20slides%20down%20the%20plane%20with%20what%0Aacceleration%3F%3C%2Fp%3E%0A%3Cp%3E%26nbsp%3B%3C%2Fp%3E%0A%3Cp%3E3)%20A%2020kg%20child%20is%20on%20a%20swing%20that%20hangs%20from%203.5%20m-long%0Achains.What%20is%20her%20maximum%20speed%20if%20she%20swings%20out%20to%20a%2045degree%0Aangle%3F%3C%2Fp%3E%0A%3Cp%3E%26nbsp%3B%3C%2Fp%3E%0A%3Cp%3E%26nbsp%3B%3C%2Fp%3E%0AExplanation / Answer
a) V^2-U^2 = 2*a*s ;
V^2-0 = 2*9.81*100 ;
V=44.29 m/s ;
b) force down the plane = mgsin30 =0.5 mg;
Max frictional force=0.3*mgcos30=0.259* mg ;
net force = 2.36*m ;
acc = 2.36 m/^2
c) max speed will be at bottom of swing ..let this to be V ;
apply work energy ;
mg(1-cos45)*3.5 = 0.5*m*v^2 ;
V=4.48 m/s
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