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Chapter 7: Problem 68

An iceboat is at rest on a frictionless frozen lake when a sudden wind exerts a constant force of \(200 \mathrm{~N}\), toward the east, on the boat. Due to the angle of the sail, the wind causes the boat to slide in a straight line for a distance of \(8.0 \mathrm{~m}\) in a direction \(20^{\circ}\) north of east. What is the kinetic energy of the iceboat at the end of that \(8.0 \mathrm{~m} ?\)

Short Answer

Expert verified
The kinetic energy of the iceboat is approximately 1503.52 joules.

Step by step solution

01

Identify the Components of the Force

Since the boat slides in a direction 20掳 north of east, we need to find the component of the force in the direction of motion. The eastward force provided by the wind is 200 N. We need to calculate the component of this force that acts in the 20掳 north of east direction using trigonometry. The component of the force can be calculated as: \[ F_{parallel} = F \cdot \cos(20^{\circ}) \]\[ F_{parallel} = 200 \cdot \cos(20^{\circ}) \]
02

Calculate Work Done

Work done by the force is calculated using the formula:\[ W = F_{parallel} \times d \]where \( F_{parallel} \) is the component of the force in the direction of motion and \( d \) is the distance. Plugging in the values:\[ W = 200 \cdot \cos(20^{\circ}) \times 8.0 \]
03

Determine the Kinetic Energy

The work-energy principle states that the work done on the object is equal to the change in kinetic energy of the object. Since the iceboat starts from rest, its initial kinetic energy is zero, and the final kinetic energy is equal to the work done by the force.\[ KE = W \]
04

Calculate the Value

Using a calculator, compute the cosine of 20掳 and multiply by the force and distance to find the work done and hence the kinetic energy:\[ F_{parallel} = 200 \cdot 0.9397 \approx 187.94 \text{ N} \]\[ W = 187.94 \times 8.0 = 1503.52 \text{ J} \]Thus, the kinetic energy of the iceboat at the end of the 8.0 meters is approximately 1503.52 joules.

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Key Concepts

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

Trigonometry in Physics
Trigonometry is a powerful tool in physics, especially when dealing with forces at angles. When a force is applied at an angle, it induces components in different directions. Imagine the boat situation here; the force is applied at a 20掳 angle north of east. This scenario tells us we have to split the 200 N force into parts for precise calculations.

The responsible trigonometric function here is the cosine, suitable for determining the force component parallel to the desired direction.
  • Calculate the parallel force using the formula: \( F_{parallel} = F \cdot \cos(\theta) \)
  • In this case, \( \theta \) is 20掳, making use of \( \cos(20掳) \)
  • This results in \( 200 \times \cos(20掳) \), which splits the 200 N force effectively for our purpose.
This concept allows you to focus calculative efforts on the actual path of motion, simplifying the physical problem by isolating force components.
Work-Energy Principle
One of the essential concepts in physics is the work-energy principle. It states the work done on an object equals its change in kinetic energy. So, when moving an object from rest, the energy exerted into motion manifests as kinetic energy. In our iceboat example, the wind performs work over an 8-meter stretch, putting energy into the boat.

Utilizing the work-energy principle involves these steps:
  • Find the component of force in the motion direction, identified earlier.
  • Work is calculated with \( W = F \times d \), where \( F \) is the force parallel to motion, and \( d \) is the distance traveled.
  • For the iceboat, substitute \( F = 187.94 \text{ N} \) and \( d = 8.0 \text{ m} \)
  • This results in work \( W = 1503.52 \text{ J} \).
With the boat initially at rest, its beginning kinetic energy is zero, so this work done equals the final kinetic energy. This principle elegantly connects force, work, and energy within a system.
Force Components
In physics, decomposing a force into components simplifies analyzing motion when forces act at angles. Consider a situation where a force isn鈥檛 perfectly aligned with the Cartesian axes could complicate calculations. Instead of attempting a direct approach:
  • Decompose force into horizontal and vertical components relative to intended paths.
  • For example, a 200 N force at 20掳 involves using the cosine for the effective moving force and sine for any perpendicular impacts.
  • This conversion provides clarity. Focus on essential directional force, ignoring non-contributing components.
Applying this skill streamlines problem-solving and is more manageable than engaging forces tangentially. Accurate force management aids in understanding overall motion and energy impacts in complex scenarios.

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Most popular questions from this chapter

A force \(\vec{F}=(2.00 \hat{1}+9.00 \hat{\mathrm{j}}+5.30 \hat{\mathrm{k}}) \mathrm{N}\) acts on a \(2.90 \mathrm{~kg}\) object that moves in time interval \(2.10 \mathrm{~s}\) from an initial position \(\vec{r}_{1}=(2.70 \hat{i}-2.90 \hat{j}+5.50 \hat{k}) \mathrm{m}\) to a final position \(\overrightarrow{\vec{r}}_{2}=\) \((-4.10 \hat{\mathrm{i}}+3.30 \hat{\mathrm{j}}+5.40 \mathrm{k}) \mathrm{m}\). Find (a) the work done on the object by the force in that time interval, (b) the average power due to the force during that time interval, and (c) the angle between vectors \(\vec{r}_{1}\) and \(\vec{r}_{2}\).

A \(250 \mathrm{~g}\) block is dropped onto a relaxed vertical spring that has a spring constant of \(k=\) \(2.5 \mathrm{~N} / \mathrm{cm}\) (Fig. \(7-46)\). The block becomes attached to the spring and compresses the spring \(12 \mathrm{~cm}\) before momentarily stopping. While the spring is being compressed, what work is done on the block by (a) the gravitational force on it and (b) the spring force? (c) What is the speed of the block just before it hits the spring? (Assume that friction is negligible.) (d) If the speed at impact is doubled, what is the maximum compression of the spring?

A helicopter lifts a \(72 \mathrm{~kg}\) astronaut \(15 \mathrm{~m}\) vertically from the ocean by means of a cable. The acceleration of the astronaut is \(g / 10 .\) How much work is done on the astronaut by (a) the force from the helicopter and (b) the gravitational force on her? Just before she reaches the helicopter, what are her (c) kinetic energy and (d) speed?

A cord is used to vertically lower an initially stationary block of mass \(M\) at a constant downward acceleration of \(g / 4\). When the block has fallen a distance \(d\), find (a) the work done by the cord's force on the block, (b) the work done by the gravitational force on the block, (c) the kinetic energy of the block, and (d) the speed of the block.

A spring with a spring constant of \(18.0 \mathrm{~N} / \mathrm{cm}\) has a cage attached to its free end. (a) How much work does the spring force do on the cage when the spring is stretched from its relaxed length by \(7.60 \mathrm{~mm} ?\) (b) How much additional work is done by the spring force when the spring is stretched by an additional \(7.60 \mathrm{~mm} ?\)
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