Q74P A skier weighing 600 N聽goes ove... [FREE SOLUTION]

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Chapter 8: Q74P (page 208)

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 $胃 = 20 掳$ . (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

$v_{a} = 6.4 \frac{m}{s}$
$v_{b \min} = 4.9 \frac{m}{s}$
If weight changes, values of $v_{a}, v_{b}$ will remain the same.

Step by step solution

Given

Weight of the skier w = 600 N
Radius of the circular hill R =20 m
Angular displacement $胃 = 20 掳$
Velocity of skier at point B $v_{b} = 8.0 \frac{m}{s}$
Frictional force and drag force are zero

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:

$K_{a} + U_{a} = K_{b} + U_{b}$

$K = \frac{1}{2} m v^{2}$

U = m g h

(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

$K_{a} + U_{a} = K_{b} + U_{b}$ (i)

At point A, $K_{a} = \frac{1}{2} m v_{a}^{2}$ (ii)

$U_{a} = m g h$ (iii)

From figure,

$\cos 胃 = \frac{R - h}{R} 鈬� \cos (20) = 1 - \frac{h}{R} 鈬� 0.9396 = 1 - \frac{h}{20}$

By solving for h, we get

h = 1.208 m (iv)

And using it in equation (ii), we get

$U_{a} m g h 鈬� U_{a} = 2 脳 9.8 脳 1.208 鈬� U_{a} = 23.6768 J K_{b} = \frac{1}{2} m v_{b}^{2} U_{b} = 0 a s B i s t h e l o w e s t p o i n t, h = 0$ (v)

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

$\frac{1}{2} m v_{a}^{2} + m g h = \frac{1}{2} m v_{b}^{2}$ .

Simplifying the equation

$v_{a}^{2} = v_{b}^{2} - 2 g h$ (vi)

$鈬� v_{a}^{2} = 40.3232 v_{a} = 6.35 \frac{m}{s} 鈮� 6.4 \frac{m}{s}$

(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

$v_{a}^{2} = v_{b}^{2} - 2 g h 鈬� 0 = v_{b}^{2} - 2 g h 鈬� v_{b}^{2} - 2 g h 鈬� v_{b}^{2} = 2 脳 9.8 脳 1.208 鈬� v_{b}^{2} = 23.6768$