1)
For the object A
applying formula "v^2=u^2+2gS"
"v^2=0+2\\times g\\times h\/2\n\\\\v=\\sqrt{gh}"
For the object B , speed at mid height
applying formula "v^2=u^2+2(-g)S"
"v^2=v_b^2-2\\times g\\times h\/2\\\\v=\\sqrt{v_b^2-gh}"
it is given that the speed of A is twice of the speed of B
"\\sqrt{gh}=2\\sqrt{v_b^2-gh}"
"gh=4(v_b^2-gh)\\\\5gh=4v_b^2\\\\h=\\dfrac{4v_b^2}{5g}"
2) Normal force = mg=10N
maximum frictional force = "\\mu"N =0.3x10=3N
so the minimum force needed to pull the book B is 3N
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