Physics
Physics
5th Edition
ISBN: 9781260486919
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 4, Problem 170P

(a)

To determine

The speed of the skier at the bottom of the slope.

(a)

Expert Solution
Check Mark

Answer to Problem 170P

The speed of the skier at the bottom of the slope is 23m/s.

Explanation of Solution

The mass of the skier is 63kg, the length of the icy slope is 50m, the angle between the slope and the horizontal is 32°, the distance covered by the skier before come to rest in horizontal path is 140m.

Diagram for skier in the slope.

Physics, Chapter 4, Problem 170P , additional homework tip  1

Write the expression to find the total force in x-direction.

Fx=maxmgsinθ=max

Here, m is the mass of the skier, ax is the acceleration of the skier, g is acceleration due to gravity, θ is the angle of the slope with horizontal.

Re-arrange the expression to find the acceleration of the skier.

ax=gsinθ

Write the expression to find the speed of the skier at the bottom of the slope.

vfx2vix2=2axΔx

Here, vfx is the final velocity in x-direction, vix is the initial velocity in x-direction, Δx is the length of the slope.

The initial velocity of the skier is zero.

Substitute 0m/s for vix to find the velocity of the skier at the bottom of the slope.

vfx20=2gsinθΔxvfx=2gsinθΔx (1)

Conclusion:

Substitute 9.8m/s2 for g, 32° for θ and 50m for Δx to find the velocity of the skier at the bottom of the slope.

vfx=2(9.8m/s2)sin32°(50m)=23m/s

(b)

To determine

The coefficient of kinetic friction between the skier and the horizontal surface.

(b)

Expert Solution
Check Mark

Answer to Problem 170P

The coefficient of kinetic friction between the skier and the horizontal surface is 0.19.

Explanation of Solution

The mass of the skier is 63kg, the length of the icy slope is 50m, the angle between the slope and the horizontal is 32°, the distance covered by the skier before come to rest in horizontal path is 140m.

Diagram for skier in horizontal surface.

Physics, Chapter 4, Problem 170P , additional homework tip  2

Write the expression to find the acceleration on the horizontal path.

vfx2vix2=2axΔxax=vfx2vix22Δx

Here, vfx is the final velocity in x-direction, vix is the initial velocity in x-direction, Δx is the length of the slope.

The final velocity of the skier is zero since he is coming to rest.

Substitute equation (1) in above expression to find the acceleration on the horizontal path.

ax=0m/s(2gsinθΔxs)22Δx=2gsinθΔxs2Δx=gsinθΔxsΔx (2)

Here, Δxs is the length of the slope.

Write the expression to find the total force in x-direction.

Fx=maxfk=max(μkN)=maxμk=maxN (3)

Here, m is the mass of the skier, ax is the acceleration of the skier, g is acceleration due to gravity.

Write the expression to find the total force in y-direction.

Fy=0Nmg=0N=mg (4)

Substitute equations (2), (4) in (3) to find the coefficient of kinetic friction between the skier and the horizontal surface.

μk=m(gsinθΔxsΔx)mg=sinθΔxsΔx

Conclusion:

Substitute 32° for θ, 140m for Δx and 50m for Δxs to find the coefficient of kinetic friction between the skier and the horizontal surface.

μk=sin32°(50m)140m=0.19

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Chapter 4 Solutions

Physics

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