/*! 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 66 A truck with mass \(m\) has a br... [FREE SOLUTION] | 91Ó°ÊÓ

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A truck with mass \(m\) has a brake failure while going down an icy mountain road of constant downward slope angle \(\alpha\) (Fig. \(\mathrm{P} 7.66\) ). 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 \(\mu_{\mathrm{r}}\) . What is the distance that the truck moves up the ramp before coming to a halt? Solve using energy methods.

Short Answer

Expert verified
The truck moves up the ramp a distance of \(d = \frac{v_0^2}{2g (\mu_r \cos \beta + \sin \beta)}\) before halting.

Step by step solution

01

Calculate Initial Energy

To determine how far the truck will move up the ramp, we start by calculating the initial mechanical energy. This consists of kinetic energy due to the truck's initial velocity (\(v_0\)) and gravitational potential energy change because of the slope angle \(\alpha\). The initial energy, where the truck starts moving downward, is solely kinetic:\[ E_{\text{initial}} = \frac{1}{2} mv_0^2 \] where \(m\) is the mass of the truck.
02

Calculate Work Done by Friction

Once the truck moves onto the ramp, friction doing work involves the force of rolling friction as the truck travels up the distance \(d\). The frictional force \(F_{\text{friction}}\) is given by:\[ F_{\text{friction}} = \mu_r mg \cos \beta \] Thus, the work done by friction, which is negative (since it opposes the movement), is:\[ W_{\text{friction}} = -\mu_r mg \cos \beta \cdot d \] where \(d\) is the distance traveled up the ramp.
03

Change in Potential Energy on the Ramp

As the truck moves up the ramp with an inclined angle \(\beta\), its gravitational potential energy increases. This can be expressed as:\[ \Delta U = mgd \sin \beta \] This represents the energy change due to height gain as the truck moves up the ramp.
04

Apply Energy Conservation

According to the law of conservation of energy, the total initial energy will be equal to the work done against friction plus the change in potential energy:\[ \frac{1}{2} mv_0^2 = \mu_r mg \cos \beta \cdot d + mgd \sin \beta \] Simplifying gives:\[ \frac{1}{2} mv_0^2 = mgd (\mu_r \cos \beta + \sin \beta) \]
05

Solve for the Distance \(d\)

Rearrange the equation from Step 4 to solve for \(d\):\[ d = \frac{\frac{1}{2} mv_0^2}{mg (\mu_r \cos \beta + \sin \beta)} \] The masses \(m\) cancel out, yielding:\[ d = \frac{v_0^2}{2g (\mu_r \cos \beta + \sin \beta)} \]This is the distance the truck will move up the ramp before it comes to a halt.

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Key Concepts

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

Kinetic Energy
Kinetic energy is the energy that an object possesses due to its motion. For an object with mass moving at velocity, the kinetic energy can be calculated using the formula:\[E_{\text{kinetic}} = \frac{1}{2} mv^2\]In the case of the truck, its initial kinetic energy is derived from its velocity \(v_0\) as it begins its journey downhill. This energy is crucial because it determines how much work is needed to stop the truck as it moves up the ramp. Since the truck's motion down the slope begins without any friction, all its energy is initially in the form of kinetic energy. This energy will eventually be transformed into other forms when the truck encounters forces that oppose its motion, such as rolling friction and the gravitational potential energy increase as it goes up the ramp.
It's important to remember that in the absence of external work, the total mechanical energy of the system is conserved. This principle allows us to connect the truck's initial kinetic energy with its energy states further along the ramp. So, understanding kinetic energy is essential for predicting how far the truck will travel up the ramp.
Gravitational Potential Energy
Gravitational potential energy is the energy stored in an object as a result of its vertical position or height relative to a lower reference point. The formula for gravitational potential energy is:\[\Delta U = mgh\]As the truck steers onto the ramp, the road slopes upwards. This incline means the truck begins to gain height, resulting in an increase in gravitational potential energy. In the context of energy conservation, the work done against gravity as the truck climbs the ramp can be expressed as follows:\[\Delta U = mgd \sin \beta\]Here, \(d\) is the distance traveled up the ramp, and \(\beta\) is the angle of the ramp. While rolling up the ramp, some of the truck's kinetic energy is converted into gravitational potential energy, enabling the truck to "climb" up the incline. Gravitational potential energy thus plays a key role in determining how far the truck can move, given that the higher the ramp, the more energy is converted from the truck's initial motion. This interplay between kinetic and potential energy is a practical demonstration of energy conservation.
Rolling Friction
Rolling friction comes into play when a round object rolls over a surface. It acts as the resistive force opposing the motion of the truck as it makes its way up the sandy ramp. This frictional force is generally much smaller than kinetic friction but significant enough to impact the truck's motion.The force resulting from rolling friction can be calculated using:\[F_{\text{friction}} = \mu_r mg \cos \beta\]Here, \(\mu_r\) is the coefficient of rolling friction, \(m\) is the mass of the truck, \(g\) is the acceleration due to gravity, and \(\beta\) is the angle of the slope. Rolling friction primarily dissipates the truck's kinetic energy into heat as the truck moves upwards, reducing the total mechanical energy available for climbing the ramp. Understanding rolling friction is crucial in this scenario, where it serves as one of the primary forces opposing the truck's movement along the ramp. By accounting for the work done against this friction, we can better predict and calculate the distance the truck will travel before it comes to a halt. Rolling friction, thus, is an essential factor in solving problems involving energy conservation in real-world scenarios.

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Most popular questions from this chapter

Cp A small block with mass 0.0500 kg slides in a vertical circle of radius \(R=0.800 \mathrm{m}\) on the inside of a circular track. There is no friction between the track and the block. At the bottom of the block's path, the normal force the track exerts on the block has magnitude 3.40 \(\mathrm{N} .\) What is the magnitude of the normal force that the track exerts on the block when it is at the top of its path?

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A \(0.150-\mathrm{kg}\) block of ice is placed against a horizontal, compressed spring mounted on a horizontal tabletop that is 1.20 \(\mathrm{m}\) above the floor. The spring has force constant 1900 \(\mathrm{N} / \mathrm{m}\) and is initially compressed 0.045 \(\mathrm{m} .\) The mass of the spring is negligible. The spring is released, and the block slides along the table, goes off the edge, and travels to the floor. If there is negligible friction between the block of ice and the tabletop, what is the speed of the block of ice when it reaches the floor?

BIO Human Energy vs. Insect Energy. For its size, the common flea is one of the most accomplished jumpers in the animal world. \(A 2.0\) -mm-long, 0.50 -mg critter can reach a height of 20 \(\mathrm{cm}\) in a single leap. (a) Neglecting air drag, what is the takeoff speed of such a flea? (b) Calculate the kinetic energy of this flea at takeoff and its kinetic energy per kilogram of mass. (c) If a \(65-\mathrm{kg}, 2.0-\mathrm{m}-\) tall human could jump to the same height compared with his length as the flea jumps compared with its length, how high could the human jump, and what takeoff speed would he need? (d) In fact, most humans can jump no more than 60 \(\mathrm{cm}\) from a crouched start. What is the kinetic energy per kilogram of mass at takeoff for such a 65-kg person? (e) Where does the flea store the energy that allows it to make such a sudden leap?

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