jack and jill are both at the top of a ski hill whose elevation is 200m. Jack sk

Question

jack and jill are both at the top of a ski hill whose elevation is 200m. Jack skies down the black diamond slope which is at an incline of 40 degree, while Jill skies down the blue slope whose inclined is 20 degrees No friction.

1) which of them arrives at the bottom at a greater speed?

2) Calculate the speed with which each arrives at the bottom of the hill?

3) which of them reaches the bottom of the hill first?

4) In what way would the answers to 1), 2) and 3) change if we include the real life case, in which there is friction?

Explanation / Answer

1) Using conservation of energy, since loss in potential energy is same for both, their velocities at the bottom will be same.

2) (1/2)mv2 = mgh

v2 = 2*9.8*200

v = 62.6 m/s

3) Acceleration for jack = gsin(40)

Acceleration for jill = gsin(20)

Since acceleration for jack is more, less time is needed to achieve same speed as compared to jill.

4) Work done by friction= F.x

= mhu.mgcos(theta).h/sin(theta)

W = mhu.mghcot(theta)

Work done by friction is more in case of jill. so jack will have higher speed.

mgh- mhu.mghcot(theta) = (1/2)mv2

v2 = 2gh-2ghcot(theta)

Related Questions

Navigate

• Browse (All)
• Browse J
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.