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Chapter 5: Problem 52

Children sled down a 41 -m-long hill inclined at \(25^{\circ} .\) At the bottom, the slope levels out. If the coefficient of friction is \(0.12,\) how far do the children slide on the level ground?

Short Answer

Expert verified
The distance the children slide on the level ground, \( d_2 \), can be calculated using gravitational potential energy, the work done against friction, and the principle of conservation of energy.

Step by step solution

01

Calculate the gravitational potential energy

When the sled with children is at top of the hill, it has a certain gravitational potential energy equals to \( m\cdot g\cdot h \) where \( m \) is mass of children and sled, \( g \) is acceleration due to gravity(9.8 m/s\(^2\)) and \( h \) is the height of the hill, which can be calculated as \( h = \sin(25^\circ) \cdot 41 \) m.
02

Calculate the total initial energy

The total initial energy of the sled with children equals to the initial gravitational potential energy only, considering zero initial kinetic energy. Hence, the initial total energy is \( m\cdot g\cdot h \).
03

Calculate the work done against friction

When the sled with children slides down, the frictional force does work against them. The work done by friction on incline and level ground can be calculated as \( W_{\text{incline}} = -\mu \cdot m \cdot g \cdot d_1 \cos(25^\circ) \) and \( W_{\text{level}} = -\mu \cdot m \cdot g \cdot d_2 \), where \( d_1 \) is length of slide down the hill and \( d_2 \) is distance slid on the level ground.
04

Apply the conservation of energy principle

The final kinetic energy of the sled with children is zero (as they come to stop eventually), and there is no change in potential energy. Hence, the initial total energy equals the work done against friction. So \( m\cdot g\cdot h=W_{\text{incline}}+W_{\text{level}}\), substituting for \(W_{\text{incline}}\) and \(W_{\text{level}}\) gives \(m\cdot g\cdot h= -\mu \cdot m \cdot g \cdot (d_1 \cos(25^\circ) + d_2))\).
05

Solve for the distance slid on the level ground

Solving the above equation for \( d_2 \) gives us the distance slid on the level ground \( d_2 = \frac{m \cdot g \cdot h +\mu \cdot m \cdot g \cdot d_1 \cos(25^\circ)}{\mu \cdot m \cdot g} \), note that \( m \) cancels out from both sides, and finally we get the distance \( d_2 \).

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Most popular questions from this chapter

A police officer investigating an accident estimates that a moving car hit a stationary car at \(25 \mathrm{km} / \mathrm{h}\). Before the collision, the car left 47 -m-long skid marks as it braked. The officer determines that the coefficient of kinetic friction was \(0.71 .\) What was the initial speed of the moving car?

In a loop-the-loop roller coaster, show that a car moving too slowly would leave the track at an angle \(\phi\) given by \(\cos \phi=v^{2} / r g\) where \(\phi\) is the angle made by a vertical line through the center of the circular track and a line from the center to the point where the car leaves the track.

A skier starts from rest at the top of a \(24^{\circ}\) slope \(1.3 \mathrm{km}\) long. Neglecting friction, how long does it take to reach the bottom?

Compare the net force on a heavy trunk when it's (a) at rest on the floor; (b) being slid across the floor at constant speed; (c) being pulled upward in an elevator whose cable tension equals the combined weight of the elevator and trunk; and (d) sliding down a frictionless ramp.

Moving through a liquid, an object of mass \(m\) experiences a resistive drag force proportional to its velocity, \(F_{\mathrm{drag}}=-b v,\) where \(b\) is a constant. (a) Find an expression for the object's speed as a function of time, when it starts from rest and falls vertically through the liquid. (b) Show that it reaches a terminal velocity \(m g / b\).
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