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A skier weighing 600 Ngoes over a frictionless circular hill of radius R = 20 m (Fig. 8-62). Assume that the effects of air resistance on the skier are negligible. As she comes up the hill, her speed is 0.8 m/s at point B, at angle θ=20°. (a) What is her speed at the hilltop (point A) if she coasts without using her poles? (b) What minimum speed can she have at B and still coast to the hilltop? (c) Do the answers to these two questions increase, decrease, or remain the same if the skier weighs 700 N instead of 600 N ?

Short Answer

Expert verified
  1. va=6.4ms
  2. vbmin=4.9ms
  3. If weight changes, values ofva,vb will remain the same.

Step by step solution

01

Given

  1. Weight of the skier w = 600 N
  2. Radius of the circular hill R =20 m
  3. Angular displacementθ=20°
  4. Velocity of skier at point Bvb=8.0ms
  5. Frictional force and drag force are zero
02

To understand the concept

If the system is isolated, only conservative forces will act on it. For this condition, we can write the sum of K and U for any state of a system = Sum of K and U for any other state of the system. Using this we can solve for unknown velocities of the system.

Formula:

Ka+Ua=Kb+Ub

K=12mv2

U = m g h

03

(a) Calculate the skier’s speed at the hilltop (point A) if she coasts without using her poles

As there is no frictional and drag force, the skier and circular hill forms the isolated system. Hence, only conservative forces are acting on the system. Therefore

Ka+Ua=Kb+Ub (i)

At point A, Ka=12mva2 (ii)

Ua=mgh (iii)

From figure,

cosθ=R-hR⇒cos20=1-hR⇒0.9396=1-h20

By solving for h, we get

h = 1.208 m (iv)

And using it in equation (ii), we get

Uamgh⇒Ua=2×9.8×1.208⇒Ua=23.6768JKb=12mvb2Ub=0asBisthelowestpoint,h=0 (v)

Using equation (ii), (iii), (iv), (v) in equation (i)

12mva2+mgh=12mvb2.

Simplifying the equation

va2=vb2-2gh (vi)

⇒va2=40.3232va=6.35ms≈6.4ms

04

(b) Calculate what minimum speed can she have at B and still coast to the hilltop 

If the skier reaches at point A with minimum velocity at point B, without using her poles, then velocity at point A will be. Using this in equation (vi), we get

va2=vb2-2gh⇒0=vb2-2gh⇒vb2-2gh⇒vb2=2×9.8×1.208⇒vb2=23.6768

Solve further as:

vb=23.6768vb=4.86ms≈4.9ms

05

(c) Find out if the answers to these two questions increase, decrease, or remain the same if the skier weighs instead of

As both equation (vi) and (vii) are independent of mass, answer forvaandvb will not change even if the weight of the skier changes.

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