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In the approximate diagram, the block with mass $m$ starts without movement in the position $P$ in the top of a hill with height $5R$. After that, the block falls down through the hill, and reaches a loop until it arrives at the position $Q$ with a height of $R$. If the loop is a circle of radius $R$ and there is no friction, what is the value of the resultant force in the point $Q$?

My try

I tried to use the law of conservation of energy in the points $P$ and $Q$, with this i get the speed, but i don't know how to continue in this problem.

Any hints?

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  • $\begingroup$ The resultant force would be the vector sum of centripetal force, and the force due to gravity $\endgroup$ – Eagle May 3 '19 at 23:05
  • $\begingroup$ @Eagle the centripetal force is in function of the mass? $\endgroup$ – Rodrigo Pizarro May 4 '19 at 0:01
  • $\begingroup$ The centripetal force is $mv^2/R$. Find v by energy conservation. $\endgroup$ – Tojrah May 4 '19 at 1:59
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Hint: the net force will be the resultant of the normal force which is horizontal (equal to centripetal force) at A and obviously the weight(=mg)which acts vertically downwards.

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  • $\begingroup$ So the sum of both gives the answer? $\endgroup$ – Rodrigo Pizarro May 4 '19 at 4:09
  • $\begingroup$ The vector sum, not simply adding as scalars! The vector sum of two perpendicular vectors $\vec a$ and $\vec b$ has magnitude $\sqrt{a^2+b^2}$ $\endgroup$ – Tojrah May 4 '19 at 4:12
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My guess is that as it's a slope we''ll divide mg into components .Mg sintheta will provide linear acceleration whereas mgcostheta will provide centripetal acceleration. Multiply by mass and u get the force .The resultant force would be the vector sum of centripetal force, and the force due to gravity.(As said in comment )

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