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Beginning from rest, an object of mass \(200 \mathrm{~kg}\) slides down a 10-m-long ramp. The ramp is inclined at an angle of \(40^{\circ}\) from the horizontal. If air resistance and friction between the object and the ramp are negligible, determine the velocity of the object, in \(\mathrm{m} / \mathrm{s}\), at the bottom of the ramp. Let \(g=9.81 \mathrm{~m} / \mathrm{s}^{2}\).

Short Answer

Expert verified
The object's velocity at the bottom of the ramp is approximately 11.23 m/s.

Step by step solution

01

Identify Given Values

The object has a mass of 200 kg and it slides down a 10 m long ramp. The angle of the ramp is 40° and the gravitational acceleration is 9.81 m/s². Initially, the object is at rest.
02

Determine the Component of Gravitational Force

Calculate the component of the gravitational force pulling the object down the ramp using: \[ F = mg \sin(\theta) \] where \( m = 200 \text{ kg} \), \( g = 9.81 \text{ m/s}^2 \), and \( \theta = 40^\circ \).
03

Calculate the Acceleration Down the Ramp

The acceleration down the ramp can be found using: \[ a = g \sin(\theta) \] where \( g = 9.81 \text{ m/s}^2 \) and \( \theta = 40^\circ \). Thus, \[ a = 9.81 \times \sin(40^\circ) \approx 6.30 \text{ m/s}^2 \].
04

Apply Kinematic Equation

Use the kinematic equation to find the final velocity: \[ v^2 = u^2 + 2as \] where the initial velocity \( u = 0 \text{ m/s} \), acceleration \( a = 6.30 \text{ m/s}^2 \), and the distance \( s = 10 \text{ m} \).
05

Solve for Final Velocity

Substitute the values into the kinematic equation: \[ v^2 = 0 + 2 \times 6.30 \text{ m/s}^2 \times 10 \text{ m} \] \[ v^2 = 126 \text{ m}^2/\text{s}^2 \] Thus, \[ v = \sqrt{126} \text{ m/s} \approx 11.23 \text{ m/s} \].

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Key Concepts

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

gravitational force
Gravitational force is a fundamental force of nature that attracts two bodies towards each other. On Earth, it gives weight to physical objects. The force of gravity acting on an object is given by the formula: \[ F = mg \]where \( F \) is the gravitational force, \( m \) is the mass of the object, and \( g \) is the acceleration due to gravity (approximated as 9.81 m/s² on Earth's surface). In our problem, this gravitational force can be broken into components because the object moves along an inclined plane. Specifically, the component of the gravitational force parallel to the ramp is responsible for accelerating the object down the slope.
acceleration
Acceleration is the rate at which an object changes its velocity. In the context of our problem, we're interested in how quickly the object increases its speed as it slides down the ramp. The equation for acceleration in this scenario is: \[ a = g \, \sin(\theta) \]where \( a \) is the acceleration, \( g \) is the gravitational acceleration (9.81 m/s²), and \( \theta \) is the angle of the slope (40°). The sine function helps us determine the portion of gravitational force that acts parallel to the ramp, causing acceleration. For our example: \[ a = 9.81 \, \sin(40^\textdegree) \approx 6.30 \text{ m/s}^2 \]
kinematic equations
Kinematic equations describe the motion of objects and are derived from Newton's laws of motion. They help us calculate key parameters like displacement, velocity, and time. One critical kinematic equation used in this problem is: \[ v^2 = u^2 + 2as \]where \( v \) is the final velocity, \( u \) is the initial velocity, \( a \) is the acceleration, and \( s \) is the displacement. Given that the object starts at rest (\( u = 0 \)), we have: \[ v^2 = 0 + 2 \times 6.30 \times 10 \approx 126 \text{ m}^2/\text{s}^2 \]By solving for \( v \), we find: \[ v = \sqrt{126} \approx 11.23 \text{ m/s} \]
inclined plane
An inclined plane is a flat surface tilted at an angle (other than a right angle) to the horizontal. It is a simple machine that allows objects to be moved up or down with less force compared to lifting them vertically. The physics of motion on an inclined plane involves breaking down forces into components parallel and perpendicular to the surface. In this exercise, the inclined plane allows us to explore concepts like component forces, acceleration due to gravity, and kinematic equations. By analyzing these components, we understand that the angle of inclination (\( \theta = 40^\textdegree \)) and the length of the ramp (10 m) significantly impact the object's acceleration and final velocity as it slides down from rest.

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Most popular questions from this chapter

For a refrigerator with automatic defrost and a topmounted freezer, the annual cost of electricity is \(\$ 55\). (a) Evaluating electricity at 8 cents per \(\mathrm{kW} \cdot \mathrm{h}\), determine the refrigerator's annual electricity requirement, in \(\mathrm{kW} \cdot \mathrm{h}\). (b) If the refrigerator's coefficient of performance is 3 , determine the amount of energy removed from its refrigerated space annually, in MJ.

A gas is contained in a vertical piston-cylinder assembly by a piston with a face area of 40 in. \(^{2}\) and weight of \(100 \mathrm{lbf}\). The atmosphere exerts a pressure of \(14.7 \mathrm{lbf} / \mathrm{in}^{2}\) on top of the piston. A paddle wheel transfers 3 Btu of energy to the gas during a process in which the elevation of the piston increases by \(1 \mathrm{ft}\). The piston and cylinder are poor thermal conductors, and friction between them can be neglected. Determine the change in internal energy of the gas, in Btu.

A gas undergoes a cycle in a piston-cylinder assembly consisting of the following three processes: Process 1-2: Constant pressure, \(p=1.4\) bar, \(V_{1}=0.028 \mathrm{~m}^{3}\), \(W_{12}=10.5 \mathrm{~kJ}\) Process 2-3: Compression with \(p V=\) constant, \(U_{3}=U_{2}\) Process 3-1: Constant volume, \(U_{1}-U_{3}=-26.4 \mathrm{~kJ}\) There are no significant changes in kinetic or potential energy. (a) Sketch the cycle on a \(p-V\) diagram. (b) Calculate the net work for the cycle, in kJ. (c) Calculate the heat transfer for process \(1-2\), in \(\mathrm{kJ}\).

A window-mounted room air conditioner removes energy by heat transfer from a room and rejects energy by heat transfer to the outside air. For steady-state operation, the air conditioner cycle requires a power input of \(0.434 \mathrm{~kW}\) and has a coefficient of performance of 6.22. Determine the rate that energy is removed from the room air, in \(\mathrm{kW}\). If electricity is valued at \(\$ 0.1 / \mathrm{kW}\) - h, determine the cost of operation for 24 hours of operation.

An object whose mass is 300 lb experiences changes in its kinetic and potential energies owing to the action of a resultant force \(\mathbf{R}\). The work done on the object by the resultant force is 140 Btu. There are no other interactions between the object and its surroundings. If the object's elevation increases by \(100 \mathrm{ft}\) and its final velocity is \(200 \mathrm{ft} / \mathrm{s}\), what is its initial velocity, in \(\mathrm{ft} / \mathrm{s}\) ? Let \(g=32.2 \mathrm{ft} / \mathrm{s}^{2}\).

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