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Chapter 5: Problem 38

A \(50 \mathrm{~kg}\) skier skis directly down a frictionless slope angled at \(10^{\circ}\) to the horizontal Assume the skier moves in the negative direction of an \(x\) axis along the slope. A wind force with component \(F_{x}\) acts on the skier. What is \(F_{x}\) if the magnitude of the skier's velocity is (a) constant, (b) increasing at a rate of \(1.0 \mathrm{~m} / \mathrm{s}^{2}\), and (c) increasing at a rate of \(2.0 \mathrm{~m} / \mathrm{s}^{2}\) ?

Short Answer

Expert verified
(a) \( F_x = -85.2 \text{ N} \), (b) \( F_x = 35.2 \text{ N} \), (c) \( F_x = -14.8 \text{ N} \).

Step by step solution

01

Identify the forces acting on the skier

The forces acting on the skier include gravity, the normal force, and the wind force. The gravitational force can be split into components along and perpendicular to the slope. The component of the gravitational force along the slope is given by \( F_{g,x} = mg \sin \theta \), where \( m = 50 \) kg is the mass of the skier, \( g = 9.8 \) m/s虏 is the acceleration due to gravity, and \( \theta = 10^\circ \) is the angle of the slope.
02

Calculate gravitational force component along the slope

Calculate the component of gravity acting along the slope: \( F_{g,x} = (50 \text{ kg})(9.8 \text{ m/s}^2)\sin(10^\circ) \). This simplifies to \( F_{g,x} \approx 85.2 \text{ N} \).
03

Evaluate condition (a) velocity constant

If the skier's velocity is constant, the net force acting along the slope must be zero. Thus, the wind force \( F_x \) must balance the gravitational force component, meaning \( F_x = -F_{g,x} \approx -85.2 \text{ N} \).
04

Evaluate condition (b) velocity increasing at 1.0 m/s虏

When the skier's velocity is increasing at a rate of 1.0 m/s虏, the net force must account for this acceleration. Apply Newton's second law, \( F_{net} = ma \), where \( a = 1.0 \text{ m/s}^2 \). Therefore, \( F_{net} = 50 \text{ kg} \times 1.0 \text{ m/s}^2 = 50 \text{ N} \). The net force is the gravitational component minus the wind force: \( F_{g,x} - F_x = 50 \text{ N} \). Solve for \( F_x \): \( F_x = 85.2 \text{ N} - 50 \text{ N} = 35.2 \text{ N} \).
05

Evaluate condition (c) velocity increasing at 2.0 m/s虏

When the skier's velocity is increasing at 2.0 m/s虏, use Newton鈥檚 second law: \( F_{net} = ma \), where \( a = 2.0 \text{ m/s}^2 \). Thus, \( F_{net} = 50 \text{ kg} \times 2.0 \text{ m/s}^2 = 100 \text{ N} \). The equation becomes \( F_{g,x} - F_x = 100 \text{ N} \). Solving for \( F_x \): \( F_x = 85.2 \text{ N} - 100 \text{ N} = -14.8 \text{ N} \).

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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 concept, playing a critical role when objects are on slopes. Imagine an invisible string pulling objects downwards - that's gravity! It always acts vertically downward towards the center of the earth. However, when considering a slope, we think about how much of this force acts along the slope. This is called the component of gravity.The formula for the component of gravitational force along a slope is: \[F_{g,x} = mg \sin \theta\]Here:
  • \(m\) is the mass of the object, in this case, the skier, which is 50 kg.
  • \(g\) is the acceleration due to gravity, approximately 9.8 m/s虏.
  • \(\theta\) is the angle of the slope, set at 10掳 in this example.
In the given problem, this force is approximately 85.2 N. It pushes the skier down the slope, playing a crucial role in determining the skier's movement.
Frictionless Motion
Frictionless motion is a scenario where we assume no resistance from the surface. Imagine sliding on a perfectly smooth surface with absolutely no slowing down - that鈥檚 what frictionless motion means! In reality, there is always some friction, but for this problem, we consider it absent for simplicity. This concept helps us focus on other forces, like gravity and wind, without additional complications. In such conditions, any motion of the skier down the slope is influenced only by:
  • Gravitational force component along the slope
  • External forces such as wind force
Without friction, there's nothing to oppose the motion initiated by gravitational pull, simplifying the calculations on how forces like wind affect movement.
Wind Force
Wind force represents additional external force acting upon the skier. It blows in the horizontal direction or along the slope, directly affecting the skier's speed.In this problem, wind force is depicted as having an effect in the same direction as the slope (either aiding or opposing gravity's pull). This force is expressed as \(F_x\) and varies depending on specific conditions:
  • When the skier's velocity is constant, wind force exactly balances the gravitational force component along the slope, so \( F_x = -85.2 \, \text{N} \).
  • When velocity increases at \(1.0 \, \text{m/s}^2\), the wind assists motion slightly less than gravity, resulting in \( F_x = 35.2 \, \text{N} \).
  • For an acceleration rate of \(2.0 \, \text{m/s}^2\), the necessary wind force is \( F_x = -14.8 \, \text{N} \) indicating it actually pushes back against motion to allow for extra speed.
Understanding wind force is crucial because it demonstrates how different accelerations require varying external influences to adjust the skier鈥檚 velocity.
Acceleration
Acceleration is all about the change in an object's velocity. If the skier's speed changes, there must be an unbalanced force causing this.Using Newton's Second Law, expressed as \( F = ma \), where:
  • \( F \) is the net external force
  • \( m \) is the mass of the object (the skier)
  • \( a \) is the acceleration
Acceleration can happen in two ways on this slope:
  • **Constant velocity**: If the velocity is constant along the slope, the forces balance out perfectly. The wind force counters everything else, meaning net acceleration is zero.
  • **Increasing velocity**: When velocity ramps up by \(1.0 \, \text{m/s}^2\), there's a needed force of 50 N to increase speed. That translates into a larger required net force when speed intensifies by \(2.0 \, \text{m/s}^2\), translating to 100 N.
In both acceleration scenarios, the changes in speed are orchestrated through a net force working alongside or against the gravitational component, thus modifying the skier's experience on the slope.

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Most popular questions from this chapter

A boy with a mass of \(35 \mathrm{~kg}\) and a sled with a mass of \(6.5 \mathrm{~kg}\) are on the frictionless ice of a frozen lake, \(12 \mathrm{~m}\) apart but connected by a rope of negligible mass. The boy exerts a horizontal \(4.2 \mathrm{~N}\) force on the rope. What are the acceleration magnitudes of (a) the sled and (b) the boy? (c) How far from the boy's initial position do they meet?

The high-speed winds around a tornado can drive projectiles into trees, building walls, and even metal traffic signs. In a laboratory simulation, a standard wood toothpick was shot by pneumatic gun into an oak branch. The toothpick's mass was \(0.13 \mathrm{~g}\), its speed before entering the branch was \(205 \mathrm{~m} / \mathrm{s}\), and its penetration depth was \(15 \mathrm{~mm}\). If its speed was decreased at a uniform rate, what was the magnitude of the force of the branch on the toothpick?

Joyce needs to lower a bundle of scrap material that weighs \(450 \mathrm{~N}\) from a point \(6.2 \mathrm{~m}\) above the ground. For doing this exercise, Joyce uses a rope that will break if the tension in it exceeds \(390 \mathrm{~N}\). Clearly if she hangs the bundle on the rope, it will break and therefore Joyce allows the bundle to accelerate downward. (a) What magnitude of the bundle's acceleration will put the rope on the verge of snapping? (b) At that acceleration, with what speed would the bundle hit the ground?

In shot putting, many athletes elect to launch the shot at an angle that is smaller than the theoretical one (about \(42^{\circ}\) ) at which the distance of a projected ball at the same speed and height is greatest. One reason has to do with the speed the athlete can give the shot during the acceleration phase of the throw. Assume that a \(7.260 \mathrm{~kg}\) shot is accelerated along a straight path of length \(1.650 \mathrm{~m}\) by a constant applied force of magnitude \(380.0 \mathrm{~N}\), starting with an initial speed of \(2.500 \mathrm{~m} / \mathrm{s}\) (due to the athlete's preliminary motion). What is the shot's speed at the end of the acceleration phase if the angle between the path and the horizontal is (a) \(30.00^{\circ}\) and (b) \(42.00^{\circ}\) ? (Hint: Treat the motion as though it were along a ramp at the given angle.) (c) By what percentage is the launch speed decreased if the athlete increases the angle from \(30.00^{\circ}\) to \(42.00^{\circ}\) ?

When two perpendicular forces \(9.0 \mathrm{~N}\) (toward positive \(x\) ) and \(7.0 \mathrm{~N}\) (toward positive \(y\) ) act on a body of mass \(6.0 \mathrm{~kg}\), what are the (a) magnitude and (b) direction of the acceleration of the body?
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