36. LetNFGbe the normal force of the ice on him and mis his mass. The net inward force ismgcos θ– FNand, according to Newton's second law, this must be equal to mv2/R,wherevis the speed of the boy. At the point where the boy leaves the ice FN= 0, so gcos = v2/R. We wish to find his speed. If the gravitational potential energy is taken to be zero when he is at the top of the ice mound, then his potential energy at the time shown is U= –mgR(1 – cos θ
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