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Chapter 5: Problem 47

A skier starts from rest at the top of a hill that is inclined \(10.5^{\circ}\) with respect to the horizontal. The hillside is \(200 \mathrm{~m}\) long, and the coefficient of friction between snow and skis is \(0.0750\). At the bottom of the hill, the snow is level and the coefficient of friction is unchanged. How far does the skier glide along the horizontal portion of the snow before coming to rest?

Short Answer

Expert verified
To find the distance the skier glides along the horizontal surface before coming to rest, first calculate the skier's velocity at the bottom of the hill considering the inclination and the friction. Then use that velocity as initial velocity to find the distance covered on the horizontal surface by considering the frictional force acting against the direction of the skier's motion.

Step by step solution

01

Calculate the skier's velocity at the bottom of the hill

Given the length of the hill \( s = 200m \), the angle of inclination \( \theta = 10.5^\circ \) and the friction coefficient \( \mu = 0.0750 \). Consider the forces acting on the skier during his descent, gravity \(mg \sin(\theta)\) acting downward along the hill and friction \(mg\cos(\theta)\mu\) acting upwards along the hill. The acceleration of the skier down the hill can be calculated using the second law of motion: \( a = g\sin(\theta) - g\cos(\theta)\mu \). Once we have the acceleration, we can find the final velocity at the bottom of the slide using the equation of motion \(v^2 = u^2 + 2as\), where \(u = 0\) (since the skier starts from rest), \(a\) is the acceleration and \(s\) is the distance. After substituting the values we get the velocity \(v\).
02

Calculate the skier's distance covered on horizontal snow

The skier then glides on level snow with the same coefficient of friction. Here the only forces acting on the skier are the force of friction, \(mg\mu\), and the force due to gravity, \(mg\). As the snow is flat, the vertical force due to gravity will be balanced out by the normal force exerted upwards by the ground. Therefore, the net force acting on the skier will be the force of friction, which acts opposite to the direction of movement. Using the second law of motion, we can find the deceleration caused due to friction \( a' = g\mu \). Now we can find the distance covered on the horizontal snow before the skier comes to rest using the equation of motion \(v^2 = u'^2 + 2a's'\), where \(u'\) is the initial velocity along the horizontal snow (which is equal to the final velocity of the descent), \(a'\) is the deceleration and \(s'\) is the distance to be covered. By substituting the respective values, we can calculate the distance\(s'\) before the skier comes to rest.
03

Putting it together

The total distance covered by the skier can be calculated by adding up the lengths of both paths; the distance covered on the hill and the distance covered on the flat snow. However, in this problem, we only need the distance covered on the flat snow which we have already calculated in the previous step.

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Most popular questions from this chapter

GP A horizontal spring attached to a wall has a force constant of \(850 \mathrm{~N} / \mathrm{m}\). A block of mass \(1.00 \mathrm{~kg}\) is attached to the spring and oscillates freely on a horizontal, frictionless surface as in Figure \(5.20\). The initial goal of this problem is to find the velocity at the equilibrium point after the block is released. (a) What objects constitute the system, and through what forces do they interact? (b) What are the two points of interest? (c) Find the energy stored in the spring when the mass is stretched \(6.00 \mathrm{~cm}\) from equilibrium and again when the mass passes through equilibrium after being released from rest. (d) Write the conservation of energy equation for this situation and solve it for the speed of the mass as it passes equilibrium. Substitute to obtain a numerical value. (c) What is the speed at the halfway point? Why isn't it half the speed at equilibrium?

A \(50.0-\mathrm{kg}\) projectile is fired at an angle of \(30.0^{\circ}\) above the horizontal with an initial speed of \(1.20 \times 10^{2} \mathrm{~m} / \mathrm{s}\) from the top of a cliff \(142 \mathrm{~m}\) above level ground, where the ground is taken to be \(y=0\). (a) What is the initial total mechanical energy of the projectile? (b) Suppose the projectile is traveling \(85.0 \mathrm{~m} / \mathrm{s}\) at its maximum height of \(y=427 \mathrm{~m}\). How much work has been done on the projectile by air friction? (c) What is the speed of the projectile immediately before it hits the ground if air friction does one and a half times as much work on the projectile when it is going down as it did when it was going up?

An archer pulls her bowstring back \(0.400 \mathrm{~m}\) by exerting a force that increases uniformly from zero to \(230 \mathrm{~N}\). (a) What is the equivalent spring constant of the bow? (b) How much work does the archer do in pulling the bow?

A truck travels uphill with constant velocity on a highway with a \(7.0^{\circ}\) slope. A \(50-\mathrm{kg}\) package sits on the floor of the back of the truck and does not slide, due to a static frictional force. During an interval in which the truck travels \(340 \mathrm{~m}\), (a) what is the net work done on the package? What is the work done on the package by (b) the force of gravity, (c) the normal force, and (d) the friction force?

W A child and a sled with a combined mass of \(50.0 \mathrm{~kg}\) slide down a frictionless slope. If the sled starts from rest and has a speed of \(3.00 \mathrm{~m} / \mathrm{s}\) at the bottom, what is the height of the hill?
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