Jack and Jill, are both at the top of a ski hill whose elevation is 200m. Jack s

Question

Jack and Jill, are both at the top of a ski hill whose elevation is 200m. Jack skis the black diamond slope which is at an incline of 40% while Jill skis down the Diu middot whose incline is 20. me there is no friction between the skis and the slopes: which of them arrives at the bottom at a greater speed. Calculate the speed with which each arrives at the bottom of the hill? which of them reaches the bottom of the hill first? In what way would the answers to a), b) and c) change if we include the real life case, in which there is friction e sure that you draw a diagram and refer to it in your answer. ohnny swings a pail of water in a vertical circle. The pail is moving (in the circle) at istant speed of 20m/s How far iS the distance from Johnny's shoulder to the floor of ail so that the water does not pour down on him, as the pail passes over his head?

Explanation / Answer

a) both are same at same height hence both possesses same potential energy that converts into KE when they come
down.

hence both will have same KE at bottom.

so speed will also be same .

b) mgh = m v^2 / 2

v = sqrt(2 gh ) = sqrt(2 x 9.8 x 200) =62.61 m/s

c) along ths incline,

F= mgsin@

so a = F/m = gsin@

and d = h/sin@

using d = ut + at^2 /2

(h / sin@) = 0 + gsin@ t^2 /2

t = sqrt(2h/ g sin^2@)

for black diamont @ is greater so sin@ will be greater hence t will be less.

Jack reahces first.

d) now work done by friction also comes in action.

f = uk N = uk mg cos@

Work done = - uk mg h (cos@ /sin@ ) = - uk mg h (cot@)

for greater @, cot@ will be less.

so work done by friction will be less.

hence will get more kE.

Jack will have more speed .

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now ,a = gsin@ - ukgcos@

h / sin@ = (gsin@ - uk g cos@) t^2 /2

a = g (sin@ - uk cos@) = g (sin40 - uk cos40)

greater tha theta, more tha acc.

hence jack will reach first.

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