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Chapter 7: Problem 58

A truck with mass \(m\) has a brake failure while going down an icy mountain road of constant downward slope angle \(\alpha\) (\(\textbf{Fig. P7.58}\)). Initially the truck is moving downhill at speed \(v_0\). After careening downhill a distance \(L\) with negligible friction, the truck driver steers the runaway vehicle onto a runaway truck ramp of constant upward slope angle \(\beta\). The truck ramp has a soft sand surface for which the coefficient of rolling friction is \(\mu_r\). What is the distance that the truck moves up the ramp before coming to a halt? Solve by energy methods.

Short Answer

Expert verified
The distance up the ramp before stopping is \(d = \frac{v_0^2 + 2gL\sin(\alpha)}{2g(\sin(\beta) + \mu_r \cos(\beta))} \).

Step by step solution

01

Analyze Initial Energy

When the truck starts, it possesses both kinetic energy due to its speed and potential energy due to its height on the slope. The initial kinetic energy is given by \( KE_i = \frac{1}{2}mv_0^2 \). Since we're using the change in height as the reference for zero potential energy, initial potential energy when the truck starts moving down the slope will depend on its vertical height change, \( PE_i = mgL\sin(\alpha) \), where \(L\sin(\alpha)\) is the vertical component of the downhill distance.
02

Analyze Energy As Truck Transitions To Ramp

As the truck reaches the base of the ramp, we calculate its speed using energy conservation on the slope. The truck has lost potential energy \(mgL\sin(\alpha)\) and maintains the kinetic energy from initial speed and loss in potential energy at the base of the ramp, calculated as \( KE_{base} = \frac{1}{2}mv_{base}^2 = \frac{1}{2}mv_0^2 + mgL\sin(\alpha) \). Using \(v_{base}\) find that \( v_{base}^2 = v_0^2 + 2gL\sin(\alpha) \).
03

Construct Energy Balance on Ramp

This balance involves accounting for the kinetic energy at the base and subtracting energy lost to friction and potential energy gained on the ramp. The ramp's vertical component affects potential energy: \( PE_{final} = mgd\sin(\beta) \). Work done against friction, \( W_f = \mu_r mgd\cos(\beta) \), reduces energy. By energy conservation: \[ \frac{1}{2}mv_{base}^2 = mgd\sin(\beta) + \mu_r mgd\cos(\beta) \].
04

Solve for Distance on Ramp

Solve the equation for distance \(d\). Start with cancelling \(m\) from the energy equation: \[ \frac{1}{2}v_{base}^2 = gd\sin(\beta) + \mu_r gd\cos(\beta) \]. Insert \(v_{base}^2 = v_0^2 + 2gL\sin(\alpha) \) from earlier, \[ \frac{1}{2}(v_0^2 + 2gL\sin(\alpha)) = gd(\sin(\beta) + \mu_r \cos(\beta)) \]. Divide and rearrange to solve for \(d\): \[ d = \frac{v_0^2 + 2gL\sin(\alpha)}{2g(\sin(\beta) + \mu_r \cos(\beta))} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kinetic Energy
In the scenario of the truck barreling down the mountain road, kinetic energy plays a significant role. Kinetic energy is essentially the energy an object possesses due to its motion, and it increases with both mass and speed.

For the truck, this initial kinetic energy (\( KE_i \)) is calculated using the formula \( KE_i = \frac{1}{2}mv_0^2 \), where \( m \) represents the mass of the truck and \( v_0 \) is its initial speed. This energy is what propels the truck forward on the icy slope.
  • Kinetic energy depends on both the mass and the square of velocity.
  • Even a slight increase in velocity results in a significant increase in kinetic energy.
  • Kinetic energy will transform as the truck's speed changes.
As the truck moves, this kinetic energy changes due to various forces it encounters, forming a crucial part of understanding energy conservation.
Potential Energy
Potential energy refers to the stored energy of an object due to its position or height. For our truck scenario, potential energy (\( PE \)) is defined relative to the height on the slope.

Initially, the truck's potential energy can be given by the expression \( PE_i = mgL\sin(\alpha) \), where \( g \) is the acceleration due to gravity, \( L \) is the distance down the slope, and \( \sin(\alpha) \) signifies the slope's angle contributing to vertical height.
  • Potential energy is maximal when the truck is at the highest initial point.
  • The formula integrates gravitational force, highlighting the impact of gravity on energy conservation.
  • The truck loses potential energy, which transforms primarily into kinetic energy as it descends.
This interplay of energies is vital for understanding how vehicles convert stored energy to motion energy, affecting how fast they travel downhill.
Rolling Friction
Rolling friction comes into play particularly when considering the truck's attempt to stop on the upward-sloping ramp. This type of friction resists the motion of a rolling object across a surface.

As the truck ascends the slope, rolling friction contributes to the energy loss that must be overcome for the truck to come to a halt. It opposes the truck鈥檚 motion up the ramp and is quantified using the coefficient of rolling friction \( \mu_r \) and the truck鈥檚 weight component aligned with the ramp鈥檚 surface: \( W_f = \mu_r mgd\cos(\beta) \).
  • Rolling friction is generally less significant than static or kinetic friction, but still critical in vehicle dynamics.
  • The angle \( \beta \) affects how gravity and rolling friction interact.
  • This resistance is key in energy calculations, influencing how far the truck can travel up the slope.
Understanding rolling friction is essential in predicting motion behavior, key for both everyday transport and energy conservation physics.

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Most popular questions from this chapter

A 2.50-kg block on a horizontal floor is attached to a horizontal spring that is initially compressed 0.0300 m. The spring has force constant 840 N/m. The coefficient of kinetic friction between the floor and the block is \(\mu_k =\) 0.40. The block and spring are released from rest, and the block slides along the floor. What is the speed of the block when it has moved a distance of 0.0200 m from its initial position? (At this point the spring is compressed 0.0100 m.)

A 75-kg roofer climbs a vertical 7.0-m ladder to the flat roof of a house. He then walks 12 m on the roof, climbs down another vertical 7.0-m ladder, and finally walks on the ground back to his starting point. How much work is done on him by gravity (a) as he climbs up; (b) as he climbs down; (c) as he walks on the roof and on the ground? (d) What is the total work done on him by gravity during this round trip? (e) On the basis of your answer to part (d), would you say that gravity is a conservative or nonconservative force? Explain.

The maximum height a typical human can jump from a crouched start is about 60 cm. By how much does the gravitational potential energy increase for a 72kg person in such a jump? Where does this energy come from?

A proton with mass \(m\) moves in one dimension. The potential-energy function is \(U(x) = (\alpha/x^2) - (\beta/x)\), where \(\alpha\) and \(\beta\) are positive constants. The proton is released from rest at \(x_0\) = \(\alpha\)/\(\beta\). (a) Show that \(U(x)\) can be written as $$U(x) = \frac{\alpha}{x^2_0}\ \Big[ \Big( \frac{x_0}{x}\ \Big)^2 - \frac{x_0}{x}\ \Big] $$ Graph \(U(x)\). Calculate \(U(x_0)\) and thereby locate the point \(x_0\) on the graph. (b) Calculate \(v(x)\), the speed of the proton as a function of position. Graph \(v(x)\) and give a qualitative description of the motion. (c) For what value of \(x\) is the speed of the proton a maximum? What is the value of that maximum speed? (d) What is the force on the proton at the point in part (c)? (e) Let the proton be released instead at \(x_1 = 3\alpha / \beta\). Locate the point \(x_1\) on the graph of \(U(x)\). Calculate \(v(x)\) and give a qualitative description of the motion. (f) For each release point (\(x = x_0\) and \(x = x_1\)), what are the maximum and minimum values of \(x\) reached during the motion?

You are an industrial engineer with a shipping company. As part of the package-handling system, a small box with mass 1.60 kg is placed against a light spring that is compressed 0.280 m. The spring has force constant \(k = 45.0\) N/m. The spring and box are released from rest, and the box travels along a horizontal surface for which the coefficient of kinetic friction with the box is \(\mu_k = 0.300\). When the box has traveled 0.280 m and the spring has reached its equilibrium length, the box loses contact with the spring. (a) What is the speed of the box at the instant when it leaves the spring? (b) What is the maximum speed of the box during its motion?
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