Answer:
13.2 m/s
Note: This answer was rounded to three significant figures.
Explanation:
To calculate the speed of the trolley after it has moved 50m down an incline, we can use the principles of energy conservation. The total mechanical energy of the system at the beginning and the end remains constant, assuming no energy is lost to the environment. However, we must account for the work done against rolling resistance.
[tex]\boxed{\begin{array}{ccc}\text{\underline{Energy Conservation with Work:}} \\\\\text{Total Energy at Start + Work Done} = \text{Total Energy at End} \\\\E_{\text{total, initial}} + W = E_{\text{total, final}} \\\\\text{Where:} \\\quad E_{\text{total, initial}} \ \text{is the total initial energy (kinetic + potential + others)} \\\quad W \ \text{is the work done on the system} \\\quad E_{\text{total, final}} \ \text{is the total final energy (kinetic + potential + others)}\end{array}}[/tex]
Given:
Here are the equations for kinetic energy, gravitational potential energy, and work:
[tex]\boxed{ \left \begin{array}{ccc} \text{\underline{Kinetic Energy:}} \\\\ K = \dfrac{1}{2}mv^2 \\\\ \text{Where:} \\ \bullet \ K \ \text{is the kinetic energy} \\ \bullet \ m \ \text{is the mass of the object} \\ \bullet \ v \ \text{is the velocity of the object} \end{array} \right.}[/tex]
[tex]\boxed{\left\begin{array}{ccc}\text{\underline{Gravitational Potential Energy: }}\\\\U_g=mgy\\\\\text{Where:}\\\bullet \ U_g\ \text{is the measure of potential energy}\\\bullet \ m\ \text{is the mass of the object}\\\bullet \ g\ \text{is the acceleration due to gravity} \ (9.8 \ m/s^2)\\\bullet \ y\ \text{is the height above a reference point}\end{array}\right}[/tex]
[tex]\boxed{\begin{array}{ccc}\text{\underline{Work Formula}:} \\\\W = Fd\cos(\theta) \\\\\text{Where:} \\\quad W \ \text{is the work done} \\\quad F \ \text{is the magnitude of the force applied} \\\quad d \ \text{is the displacement of the object} \\\quad \theta \ \text{is the angle between the force and the displacement vector}\end{array}}[/tex]
Using the attached image for reference, we can form the following equation using points (1) and (2):
[tex]U_{g_1}-W_{\vec F_r}=K_2[/tex]
The trolley starts from rest at point (1) and travels 50 meters to point (2), the resistive force does negative work, as it is dissipating energy (cos(180°) = -1). Let's expand the above equation:
[tex]\Longrightarrow mgy-\vec F_rd=\dfrac{1}{2}mv^2[/tex]
From the above equation, our unknown quantity is 'v'. We also do not know 'y' but can find it using the sine function:
[tex]\Longrightarrow mgd\sin(\theta)-\vec F_rd=\dfrac{1}{2}mv^2[/tex]
Now we can plug in our know values and solve for 'v':
[tex]\Longrightarrow (500 \text{ kg})(9.8 \text{ m/s}^2)(50 \text{ m})\sin(20^\circ)- (800 \text{ N})(50 \text{ m})=\dfrac{1}{2}(500 \text{ kg})v^2[/tex]
[tex]\Longrightarrow 83794.9 \text{ J}- 40000 \text{ J}=(250 \text{ kg})v^2[/tex]
[tex]\Longrightarrow 43794.9 \text{ J}=(250 \text{ kg})v^2\\\\\\\\[/tex]
[tex]\Longrightarrow v=\sqrt{\dfrac{43794.9 \text{ J}}{250 \text{ kg}}}[/tex]
[tex]\therefore v \approx \boxed{13.2 \text{ m/s}}[/tex]
Thus, the speed of the trolley after moving 50m down the incline is approximately 13.2 m/s.
