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A motorcycle and \(60.0 \mathrm{~kg}\) rider accelerate at \(3.0 \mathrm{~m} / \mathrm{s}^{2}\) up a ramp inclined \(10^{\circ}\) above the horizontal. What are the magnitudes of (a) the net force on the rider and (b) the force on the rider from the motorcycle?

Short Answer

Expert verified
(a) 180.0 N net force; (b) motorcycle exerts a larger force than net, considering incline friction.

Step by step solution

01

Calculate the Net Force on the Rider

To find the net force, use Newton's second law of motion. The formula is \( F = ma \), where \( m \) is the mass of the rider and \( a \) is the acceleration.Given:\[ m = 60.0 \text{ kg} \] \[ a = 3.0 \text{ m/s}^2 \]Calculate:\[ F = 60.0 \text{ kg} \times 3.0 \text{ m/s}^2 = 180.0 \text{ N} \]Therefore, the net force on the rider is \( 180.0 \text{ N} \).
02

Determine the Components of Forces Acting on the Rider

Identify forces acting on the rider. These include gravity acting down the ramp and the normal force perpendicular to the ramp. The force due to gravity parallel to the ramp can be calculated by:\[ F_{g, parallel} = mg \sin(\theta) \]Where:- \( g = 9.8 \text{ m/s}^2 \) is the acceleration due to gravity.- \( \theta = 10^{\circ} \) is the angle of the incline.Calculate:\[ F_{g, parallel} = 60.0 \text{ kg} \times 9.8 \text{ m/s}^2 \times \sin(10^{\circ}) \approx 102.3 \text{ N} \].

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Key Concepts

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

Net Force Calculation
When assessing the motion of any object, calculating the net force is a fundamental step. Newton's Second Law of Motion provides the perfect tool for this task. It states that the net force (\( F \)) acting on an object equals the mass (\( m \)) of the object times its acceleration (\( a \)):
  • Formula: \( F = ma \)

For the motorcycle rider, we started by identifying the given values:
  • Mass (\( m \)) = 60.0 kg
  • Acceleration (\( a \)) = 3.0 m/s²
Substituting these into the formula, we computed the net force:\[ F = 60.0 \, \text{kg} \times 3.0 \, \text{m/s}^2 = 180.0 \, \text{N} \]
This tells us that a force of 180 N is required to accelerate the rider up the inclined plane. The net force encompasses all acting forces, including gravitational components and the potential force from the motorcycle itself.
Force Components
On an inclined plane, various forces can impact an object. It's essential to break these down to understand their contributions. For the motorcycle scenario, consider these primary force components:
  • Gravitational force acting downwards.
  • Normal force perpendicular to the slope.
  • Force parallel to the incline due to gravity.

The gravitational force parallel to the slope is crucial because it opposes the upward acceleration:\[ F_{g, parallel} = mg \sin(\theta) \]
Where \( g = 9.8 \, \text{m/s}^2 \) is gravity's acceleration, and \( \theta = 10^\circ \) is the ramp's angle. Calculate this component:
\[ F_{g, parallel} = 60.0 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times \sin(10^\circ) \approx 102.3 \, \text{N} \]
This force works against the rider's motion, indicating how much of the motorcycle's force must counteract gravity. Understanding these components helps us appreciate the physics behind inclined plane dynamics and ensures we apply the right amount of force for desired motion.
Inclined Plane Dynamics
Inclined planes are commonplace in physics problems because they introduce concepts like force components and gravitational effects at play. Understanding their dynamics is key.
For the rider on the inclined ramp, these concepts play out as follows:
  • The net force moves the rider uphill, counteracting the sloped gravitational force.
  • Gravitational forces can be divided into components: parallel (affects movement) and perpendicular (affects normal force).
  • The normal force is perpendicular to the ramp's surface and balances the gravitational pull.

As you study inclined plane dynamics, remember:
  • It's important to calculate each force component accurately.
  • In real-world applications, friction might also play a vital role, although it's not a focus here.

By dividing forces into their components, such problems become more manageable, helping us analyze objects in motion on slopes accurately.

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Most popular questions from this chapter

An \(80 \mathrm{~kg}\) person is parachuting and experiencing a downward acceleration of \(2.5 \mathrm{~m} / \mathrm{s}^{2}\). The mass of the parachute is \(5.0 \mathrm{~kg}\). (a) What is the upward force on the open parachute from the air? (b) What is the downward force on the parachute from the person?

An object is hung from a spring balance attached to the ceiling of an elevator cab. The balance reads \(65 \mathrm{~N}\) when the cab is standing still. What is the reading when the cab is moving upward (a) with a constant speed of \(7.6 \mathrm{~m} / \mathrm{s}\) and (b) with a speed of \(7.6 \mathrm{~m} / \mathrm{s}\) while decelerating at a rate of \(2.4 \mathrm{~m} / \mathrm{s}^{2} ?\)

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The tension at which a fishing line snaps is commonly called the line's "strength." What minimum strength is needed for a line that is to stop a salmon of weight \(85 \mathrm{~N}\) in \(11 \mathrm{~cm}\) if the fish is initially drifting at \(2.8 \mathrm{~m} / \mathrm{s}\) ? Assume a constant deceleration.

A hot-air balloon of mass \(M\) is descending vertically with downward acceleration of magnitude \(a\). How much mass (ballast) must be thrown out to give the balloon an upward acceleration of magnitude \(a\) ? Assume that the upward force from the air (the lift) does not change because of the decrease in mass.

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