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Chapter 4: Problem 69

A student is skateboarding down a ramp that is \(6.0 \mathrm{~m}\) long and inclined at 18 with respect to the horizontal. The initial speed of the skateboarder at the top of the ramp is \(2.6 \mathrm{~m} / \mathrm{s}\). Neglect friction and find the speed at the bottom of the ramp.

Short Answer

Expert verified
The skateboarder's speed at the bottom of the ramp is approximately \(8.5 \, \mathrm{m/s}\).

Step by step solution

01

Identify the Known Quantities

We know that the ramp length is \(6.0\, \mathrm{m}\), the initial speed of the skateboarder \(v_i = 2.6\, \mathrm{m/s}\), and the angle of inclination \(\theta = 18^\circ\). We also know acceleration due to gravity is \(g = 9.8\, \mathrm{m/s}^2\).
02

Find the Component of Gravitational Acceleration

The acceleration along the ramp due to gravity can be found using the formula \(a = g \cdot \sin(\theta)\). This gives: \[a = 9.8 \cdot \sin(18^\circ)\].
03

Use the Kinematic Equation

We can use the kinematic equation \(v_f^2 = v_i^2 + 2a \cdot d\), where \(v_f\) is the final speed, \(v_i\) the initial speed, \(a\) the acceleration, and \(d\) the distance. Substitute the known values, including \(a\) from Step 2 and \(d = 6.0\, \mathrm{m}\).
04

Calculate the Final Speed

Substitute the values into the kinematic equation: \[v_f^2 = (2.6)^2 + 2 \cdot a \cdot 6.0\]. Solve for \(v_f\) after calculating \(a\) from the expression \(a = 9.8 \cdot \sin(18^\circ)\).
05

Compute and Conclude

Upon solving the equation in Step 4, we find \(v_f\). Plug in all calculated values to finish finding the final speed.

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Key Concepts

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

Gravitational Acceleration
Gravitational acceleration is the force that draws objects toward the center of the Earth, denoted by the constant \( g = 9.8 \space \mathrm{m/s}^2 \). This universal force acts on every object in free fall, leading them to accelerate downward. In contexts like this skateboard ramp scenario, gravitational force plays a crucial role in influencing motion.
  • When an object moves down an inclined plane, only a portion of gravitational acceleration acts along the incline.
  • This relevant portion is computed as \( a = g \cdot \sin(\theta) \), where \( \theta \) is the angle of the incline relative to the horizontal.
Understanding how gravity affects motion on an incline helps us calculate changes in speed and predict future states of movement.
Inclined Plane
An inclined plane is a simple machine that allows objects to be moved between different heights more easily. In an ideal case, when friction is neglected, predicting motion becomes more straightforward.
  • The ramp's angle, \( \theta = 18^\circ \), determines how much the gravitational force contributes to accelerating the skateboarder down the ramp.
  • The longer the inclined plane, in this case, \( 6.0 \space \mathrm{m} \), the more distance there is for the force to act, influencing the speed increase.
Inclined planes are a frequent focus in physics problems because they offer a practical way to explore gravitational forces and acceleration in a controlled manner.
Kinematic Equations
Kinematic equations are essential tools for analyzing motion, allowing us to connect quantities like displacement, velocity, and acceleration. In this ramp example, we leverage these equations to predict the final speed of the skateboarder.
  • One key equation is \( v_f^2 = v_i^2 + 2a \cdot d \), where \( v_f \) is the final velocity, \( v_i \) the initial velocity, \( a \) the acceleration, and \( d \) the distance traveled.
  • This formula helps determine the velocity at any point, given known initial conditions and distance.
By inserting known values like initial speed, ramp length, and calculated acceleration, we can solve for the final speed at the bottom. Kinematic equations provide a systematic method to unlock insights about motion in physics.

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Most popular questions from this chapter

A toboggan slides down a hill and has a constant velocity. The angle of the hill is \(8.00^{\circ}\) with respect to the horizontal. What is the coefficient of kinetic friction between the surface of the hill and the toboggan?

When a 58 -g tennis ball is served, it accelerates from rest to a speed of \(45 \mathrm{~m} / \mathrm{s}\). The impact with the racket gives the ball a constant acceleration over a distance of \(44 \mathrm{~cm} .\) What is the magnitude of the net force acting on the ball?

A person with a black belt in karate has a fist that has a mass of \(0.70 \mathrm{~kg}\). Starting from rest, this fist attains a velocity of \(8.0 \mathrm{~m} / \mathrm{s}\) in \(0.15 \mathrm{~s}\). What is the magnitude of the average net force applied to the fist to achieve this level of performance?

A car is towing a boat on a trailer. The driver starts from rest and accelerates to a velocity of \(+11 \mathrm{~m} / \mathrm{s}\) in a time of \(28 \mathrm{~s}\). The combined mass of the boat and trailer is 410 kg. The frictional force acting on the trailer can be ignored. What is the tension in the hitch that connects the trailer to the car?

A block whose weight is \(45.0 \mathrm{~N}\) rests on a horizontal table. A horizontal force of \(36.0 \mathrm{~N}\) is applied to the block. The coefficients of static and kinetic friction are 0.650 and 0.420 , respectively. Will the block move under the influence of the force, and, if so, what will be the block's acceleration? Explain your reasoning.
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