Skip to main content
Ch. 6 - Applications of Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243당신이 사용하는 게 아니라요?교과서 변경
6장, 문제 6.7.57c

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.
c. The work required to lift a 10-kg object vertically 10 m is the same as the work required to lift a 20-kg object vertically 5 m.

검증된 단계별 안내
1
Recall the formula for work done against gravity when lifting an object vertically: \(W = m \cdot g \cdot h\), where \(m\) is the mass of the object, \(g\) is the acceleration due to gravity (approximately \(9.8\, m/s^2\)), and \(h\) is the height the object is lifted.
Calculate the work done to lift the 10-kg object 10 meters: \(W_1 = 10 \cdot g \cdot 10\).
Calculate the work done to lift the 20-kg object 5 meters: \(W_2 = 20 \cdot g \cdot 5\).
Compare \(W_1\) and \(W_2\) by simplifying both expressions to see if they are equal or not.
Conclude whether the statement is true or false based on the comparison, and explain that work depends on both mass and height, so equal work means the product \(m \cdot h\) must be the same.

비슷한 문제에 대한 검증된 영상 답변:

이 영상 해법은 위 문제에 도움이 된다고 튜터들이 추천한 것입니다.
영상 길이:
2m
도움이 되었나요?

주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Work Done by a Force

Work is defined as the product of the force applied to an object and the displacement in the direction of the force. Mathematically, work = force × distance × cos(θ). For lifting vertically, the force equals the weight of the object, and displacement is the vertical height moved.
추천 영상:
가이드 코스
05:40
Work Done On A Spring (Hooke's Law)

Weight and Gravitational Force

Weight is the gravitational force acting on an object and is calculated as mass times gravitational acceleration (W = mg). Different masses result in different weights, which directly affect the amount of work needed to lift the object.
추천 영상:
가이드 코스
09:32
Lifting Problems

Comparing Work for Different Masses and Distances

To compare work done lifting different masses over different heights, multiply each mass by gravitational acceleration and the height lifted. If the products are equal, the work done is the same; otherwise, it differs. This helps determine if lifting a heavier object a shorter distance equals lifting a lighter object a longer distance.
추천 영상:
가이드 코스
09:32
Lifting Problems
관련 실천
교과서 질문

Oscillating growth rates Some species have growth rates that oscillate with an (approximately) constant period P. Consider the growth rate function N'(t) = r+A sin 2πt/P, where A and r are constants with units of individuals/yr, and t is measured in years. A species becomes extinct if its population ever reaches 0 after t=0.


c. Suppose P=10, A=50, and r=5. If the initial population is N(0)=10, does the population ever become extinct? Explain.

42
views
교과서 질문

Region R is revolved about the line y=1 to form a solid of revolution.


c. Write an integral for the volume of the solid.

73
views
교과서 질문

Let R be the region bounded by the curve y=√cos x and the x-axis on [0, π/2]. A solid of revolution is obtained by revolving R about the x-axis (see figures). 


c. Write an integral for the volume of the solid.

64
views
교과서 질문

13–16. Displacement from velocity Consider an object moving along a line with the given velocity v. Assume time t is measured in seconds and velocities have units of m/s.


c. Find the distance traveled over the given interval.


v(t) = 4t³ - 24t²+20t on [0, 5]

67
views
교과서 질문

Acceleration A drag racer accelerates at a(t)=88 ft/s². Assume v(0)=0, s(0)=0, and t is measured in seconds.


c. At this rate, how long will it take the racer to travel 1/4 mi?

46
views
교과서 질문

9–10. Velocity graphs The figures show velocity functions for motion along a line. Assume the motion begins with an initial position of s(0)=0. Determine the following.

c. The position at t=5

40
views