/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 47 A skier starts from rest at the ... [FREE SOLUTION] | 91影视

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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

In the dangerous "sport" of bungee jumping, a daring student jumps from a hot- air balloon with a specially designed elastic cord attached to his waist. The unstretched length of the cord is \(25.0 \mathrm{~m}\), the student weighs \(700 \mathrm{~N}\), and the balloon is \(36.0 \mathrm{~m}\) above the surface of a river below. Calculate the required force constant of the cord if the student is to stop safely \(4.00 \mathrm{~m}\) above the river.

\(\mathrm{S}\) A projectile of mass \(m\) is fired horizontally with an initial speed of \(v_{0}\) from a height of \(h\) above a flat, desert surface. Neglecting air friction, at the instant before the projectile hits the ground, find the following in terms of \(m, v_{0}, h\), and \(g\) : (a) the work done by the force of gravity on the projectile, (b) the change in kinetic energy of the projectile since it was fired, and (c) the final kinetic energy of the projectile. (d) Are any of the answers changed if the initial angle is changed?

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?

When an automobile moves with constant speed down a highway, most of the power developed by the engine is used to compensate for the mechanical energy loss due to frictional forces exerted on the car by the air and the road. If the power developed by an engine is 175 hp, estimate the total frictional force acting on the car when it is moving at a speed of \(29 \mathrm{~m} / \mathrm{s}\). One horsepower cquals \(746 \mathrm{~W}\).

A child's pogo stick (Fig. P5.81) stores energy in a spring \((k=\) \(\left.2.50 \times 10^{4} \mathrm{~N} / \mathrm{m}\right)\). At position (A) \(\left(x_{1}=-0.100 \mathrm{~m}\right)\), the spring compression is a maximum and the child is momentarily at rest. At position (B) \((x=0)\), the spring is relaxed and the child is moving upward. At position (C), the child is again momentarily at rest at the top of the jump. Assuming that Figure p \(5.81\) the combined mass of child and pogo stick is \(25.0 \mathrm{~kg}\), (a) calculate the total energy of the system if both potential energies are zero at \(x=0\), (b) determine \(x_{2}\), (c) calculate the speed of the child at \(x=0\), (d) determine the value of \(x\) for which the kinetic energy of the system is a maximum, and (c) obtain the child's maximum upward speed.
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