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(II) A skateboarder, with an initial speed of 2.0 m/s rolls virtually friction free down a straight incline of length 18 m in 3.3 s. At what angle \(\theta\) is the incline oriented above the horizontal?

Short Answer

Expert verified
The incline is oriented at approximately \( 12.3^\circ \) above the horizontal.

Step by step solution

01

Understanding the Problem

We need to find the angle \( \theta \) at which an incline is oriented, knowing the skateboarder's initial speed, the distance of the incline, and the time taken to travel down.
02

Define Given Values

The given values are \( u = 2.0 \text{ m/s} \) (initial speed), \( s = 18 \text{ m} \) (distance), and \( t = 3.3 \text{ s} \) (time).
03

Calculate the Final Speed

We suspect the problem involves uniformly accelerated motion without friction, so we first calculate the final speed using the equation \( v = u + at \). However, we need \( v \) to find acceleration using \( s = \frac{u + v}{2} \cdot t \).
04

Calculate the Average Velocity

Using the equation \( s = \frac{u + v}{2} \cdot t \), solve for \( v \). Substituting the known values: \[ 18 = \frac{2 + v}{2} \times 3.3 \] This gives \( v = 8.909 \text{ m/s} \).
05

Determine Acceleration

Substitute \( v = 8.909 \text{ m/s} \) into \( v = u + at \) to find \( a \). Rearranging gives: \[ a = \frac{v - u}{t} = \frac{8.909 - 2.0}{3.3} = 2.094 \text{ m/s}^2 \]
06

Relate Acceleration to Gravity

The formula for acceleration along an incline is \( a = g \sin(\theta) \), where \( g \approx 9.8 \text{ m/s}^2 \) is the acceleration due to gravity. Solving for \( \theta \): \[ \sin(\theta) = \frac{a}{g} = \frac{2.094}{9.8} \approx 0.213 \]
07

Calculate the Angle \( \theta \)

Use the inverse sine function to determine the angle: \( \theta = \arcsin(0.213) \approx 12.3^\circ \).

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Key Concepts

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

Initial Speed
Initial speed is a fundamental parameter in motion problems, particularly those concerning uniformly accelerated motion. It refers to the speed of an object at the moment it starts moving. In the context of the skateboarder rolling down an incline, knowing the initial speed helps in figuring out the dynamics of the motion.
This speed is denoted as \( u \) in equations and is critical in calculations involving velocity and acceleration. The initial speed is part of the primary motion equations that help predict and describe an object's motion.
In scenarios like the skateboarder problem, the given initial speed was 2.0 m/s. This means at the start of the incline, the skateboarder is moving forward at a speed of 2.0 m/s. From this initial speed, we can determine how quickly the skateboarder accelerates as they travel down the incline.
Inclined Plane
An inclined plane is a flat surface tilted at an angle, other than zero, to the horizontal. It's a classic example found in physics to investigate concepts of force and motion. The skateboarder scenario uses an inclined plane to explore motion dynamics downhill.
The angle of the incline, denoted as \( \theta \), significantly affects how objects accelerate down the plane. In principle, the steeper the angle, the greater the acceleration, assuming other forces such as friction are minimal. In our example, the problem asks for the angle of the incline based on the skateboarder's motion parameters.
Inclined planes allow us to explore gravitational force components, breaking them down into parallel and perpendicular forces. The skateboarder's movement down the incline is influenced primarily by the force component along the plane, which decides the acceleration, impacting speed and travel time.
Acceleration Calculation
Acceleration is the rate at which an object's velocity changes over time and is a central idea in uniformly accelerated motion problems. In our skateboarder problem, acceleration is paramount to understanding how fast the skateboarder speeds up as they move down the incline.
Given the initial speed, final speed, and time, we use the equation \( v = u + at \) to find acceleration \( a \). Here, \( a \) is calculated by rearranging the formula to \( a = \frac{v-u}{t} \). This equation links initial speed, final speed, and time with acceleration.
  • The final calculated speed of the skateboarder helps ascertain the uniform acceleration experienced.
  • By further linking this acceleration to gravitational acceleration \( g \), through the formula \( a = g \sin(\theta) \), we relate the physical inclination to theoretical parameters.
    This provides a logical method to calculate the incline angle. Such calculations offer practical insights in fields like engineering and physics where inclined planes are frequently encountered.

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Most popular questions from this chapter

Piles of snow on slippery roofs can become dangerous projectiles as they melt. Consider a chunk of snow at the ridge of a roof with a slope of 34\(^\circ\). (\(a\)) What is the minimum value of the coefficient of static friction that will keep the snow from sliding down? (\(b\)) As the snow begins to melt, the coefficient of static friction decreases and the snow finally slips. Assuming that the distance from the chunk to the edge of the roof is 4.0 m and the coefficient of kinetic friction is 0.10, calculate the speed of the snow chunk when it slides off the roof. (\(c\)) If the roof edge is 10.0 m above ground, estimate the speed of the snow when it hits the ground.

(III) A child slides down a slide with a 34\(^\circ\) incline, and at the bottom her speed is precisely half what it would have been if the slide had been frictionless. Calculate the coefficient of kinetic friction between the slide and the child.

Two blocks made of different materials, connected by a thin cord, slide down a plane ramp inclined at an angle to the horizontal, Fig. 4-76 (block \(B\) is above block \(A\)). The masses of the blocks are \(m_A\) and \(m_B\) and the coefficients of friction are \(\mu_A\) and \(\mu_B\). If \(m_A=m_B=\) 5.0 kg and \(\mu_A=\) 0.20 and \(\mu_B=\) 0.30, determine (a) the acceleration of the blocks and (b) the tension in the cord, for an angle \(\theta =\) 32\(^\circ\).

(II) A car can decelerate at \(-\)3.80 m/s\(^2\) without skidding when coming to rest on a level road. What would its deceleration be if the road is inclined at 9.3\(^\circ\) and the car moves uphill? Assume the same static friction coefficient.

A 75.0-kg person stands on a scale in an elevator. What does the scale read (in N and in kg) when (\(a\)) the elevator is at rest, (\(b\)) the elevator is climbing at a constant speed of 3.0 m/s, (\(c\)) the elevator is descending at 3.0 m/s, (\(d\)) the elevator is accelerating upward at 3.0 m/s\(^2\), (\(e\)) the elevator is accelerating downward at 3.0 m/s\(^2\)?

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