A 60 000-kg train is being dragged along a straight line up...

A 60 000-kg train is being dragged along a straight line up a $1.0$ percent grade (i.e., the road rises $1.0 \mathrm{~m}$ for each $100 \mathrm{~m}$ traveled horizontally) by a steady drawbar pull of $3.0 \mathrm{kN}$ parallel to the incline. The friction force opposing the motion of the train is $4.0$ $\mathrm{kN}$. The train's initial speed is $12 \mathrm{~m} / \mathrm{s}$. Through what distance s will the train move along its tracks before its speed is reduced to $9.0 \mathrm{~m} / \mathrm{s}$ ? Use energy considerations. The change in total energy of the train is due to the work done by the friction force (which is negative) and the drawbar pull (which is positive): Change in $\mathrm{KE}+$ change in $\mathrm{PE}_{\mathrm{G}}=W_{\text {drawbar }}+W_{\text {friction }}$ The train loses $\mathrm{KE}$ and gains $\mathrm{PE}_{\mathrm{G}}$. It rises a height $h=s \sin \theta$, where $\theta$ is the incline angle and $\tan \theta=1 / 100$. Hence, $\theta=0.573^{\circ}$, and $h=0.010 \mathrm{~s}$ (at small angles $\tan \theta \approx \sin \theta$ ). Therefore, $$ \begin{array}{c} \frac{1}{2} m\left(v_{f}^{2}-v_{i}^{2}\right)+m g(0.010 s)=(3000 \mathrm{~N})(s)(1)+(4000 \mathrm{~N})(s)(-1) \\ -1.89 \times 10^{6} \mathrm{~J}+\left(5.89 \times 10^{3} \mathrm{~N}\right) s=(-1000 \mathrm{~N}) s \end{array} $$ from which we obtain $s=274 \mathrm{~m}=0.27 \mathrm{~km}$.

A 60 000-kg train is being dragged along a straight line up a $1.0$ percent grade (i.e., the road rises $1.0 \mathrm{~m}$ for each $100 \mathrm{~m}$ traveled horizontally) by a steady drawbar pull of $3.0 \mathrm{kN}$ parallel to the incline. The friction force opposing the motion of the train is $4.0$ $\mathrm{kN}$. The train's initial speed is $12 \mathrm{~m} / \mathrm{s}$. Through what distance s will the train move along its tracks before its speed is reduced to $9.0 \mathrm{~m} / \mathrm{s}$ ? Use energy considerations.
The change in total energy of the train is due to the work done by the friction force (which is negative) and the drawbar pull (which is positive):
Change in $\mathrm{KE}+$ change in $\mathrm{PE}_{\mathrm{G}}=W_{\text {drawbar }}+W_{\text {friction }}$
The train loses $\mathrm{KE}$ and gains $\mathrm{PE}_{\mathrm{G}}$. It rises a height $h=s \sin \theta$, where $\theta$ is the incline angle and $\tan \theta=1 / 100$. Hence, $\theta=0.573^{\circ}$, and $h=0.010 \mathrm{~s}$ (at small angles $\tan \theta \approx \sin \theta$ ). Therefore,
$$
\begin{array}{c}
\frac{1}{2} m\left(v_{f}^{2}-v_{i}^{2}\right)+m g(0.010 s)=(3000 \mathrm{~N})(s)(1)+(4000 \mathrm{~N})(s)(-1) \\
-1.89 \times 10^{6} \mathrm{~J}+\left(5.89 \times 10^{3} \mathrm{~N}\right) s=(-1000 \mathrm{~N}) s
\end{array}
$$ from which we obtain $s=274 \mathrm{~m}=0.27 \mathrm{~km}$.

Key Concepts

Small Angle Approximation on Inclines

When dealing with inclined planes that have small angles, the small angle approximation is often used, where the sine of the angle is approximated by the tangent of the angle. This simplification facilitates the analysis of vertical height gained over a distance along the incline, thereby easing the calculation of gravitational potential energy changes.

Work Done by Forces

The work done by forces on an object is calculated as the force multiplied by the displacement in the direction of the force. This encompasses both positive work, which adds energy to the system, and negative work, which removes energy. Understanding how individual forces such as applied forces or friction contribute to the net work is key in energy-based problem solving.

Gravitational Potential Energy

Gravitational potential energy is the energy stored in an object because of its position in a gravitational field, typically calculated as the product of its mass, gravitational acceleration, and height above a reference level. It is essential for understanding energy changes when an object moves vertically, such as when ascending or descending an incline.

Kinetic Energy

Kinetic energy is the energy an object has due to its motion, defined as one half the mass times the square of the velocity. It is a fundamental concept for energy analysis in mechanics, as changes in kinetic energy indicate acceleration or deceleration resulting from work done by forces.

Work-Energy Principle

The work-energy principle states that the net work done on an object is equal to the change in its kinetic energy. This concept is central to analyzing systems where forces (both conservative and non-conservative) influence the motion of an object, allowing one to relate the effects of these forces directly to changes in speed or kinetic energy.

Transcript

0:00 Hello.

00:02 In this question, we have a train that is being dragged on an incline, and for this incline, we're given the following information.

00:11 For each 100 meters traveled horizontally, the level of the incline increases by one meter.

00:21 So this is a fact about the inclination itself.

00:25 Now, we know that that weight of the train is 60 ,000 kids.

00:31 And this train is being dragged as we said by a force that is equal to three kiloons and the friction between the train and rails is four kiloons so the question here given that the train was moving with an initial velocity of 12 meters per second up or up decline we want to know how far or what the distance on the plane will the train be but if its velocity is now 9 meters per second.

01:11 So that's the question.

01:13 And the question can be solved by just mentioning the equation of the change in energy and its equality with the work done on the object.

01:23 So here, the change in the kinetic energy of the train plus the change in the potential energy of the train is equal to the total work done on the train.

01:33 And the work done is by two forces here.

01:36 First, the work done by the force of the drag that is dragging the train up the climb and the work done by the friction that is mainly dragging or that is mainly opposing the motion of the train or doing negative work on the train.

01:59 So let's start by substituting values here.

02:03 So we have the change in the kinetic energy is half multiplied by the mass of the train, multiplied by the difference between the squares of velocities, the final velocity and the initial velocity.

02:17 The change in potential energy is just the mass multiplied by the gravitational acceleration, multiplied by the change in height or the change in altitude of the train.

02:28 And the change in height in that case, if we make a triangle here, so if that's distance d, so the change in height, this would be equal to d sine theta, where theta is the angle of the inclination here...

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