For Exercises 65-68, find the work w done by a force F in moving an object in a straight line given by the displacement vector D. (See Example 6) F = − 26 i + 32 j N ; D = 100 i + 120 j m
For Exercises 65-68, find the work w done by a force F in moving an object in a straight line given by the displacement vector D. (See Example 6) F = − 26 i + 32 j N ; D = 100 i + 120 j m
Solution Summary: The author calculates the work done by an external force, F=(-26i+32j)N, to move an object in a straight line for the given displacement vector,
Find the work done by a force of 200 pounds acting in the direction -1+2j in moving
an object 75 feet from (0, 0) to (-75, 0).
15,000.0 ft-lb
13,416.1 ft-lb
6708.2 ft-lb
8944.9 ft-lb
You are on a rollercoaster, and the path of your body is modeled by a vector function r(t),
where t is in seconds, the units of distance are in feet, and t = 0 represents the start of the
ride. Assume the axes represent the standard cardinal directions and elevation (x is E/W, y
is N/S, z is height). Explain what the following would represent physically, being as specific
as possible. These are all common roller coaster shapes/behaviors and can be explained in
specific language with regard to units:
a. r(0)=r(120)
b. For 0 ≤ t ≤ 30, N(t) = 0
c. r'(30) = 120
d. For 60 ≤ t ≤ 64, k(t) =
40
and z is constant.
e.
For 100 ≤ t ≤ 102, your B begins by pointing toward positive z, and does one full
rotation in the normal (NB) plane while your T remains constant.
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Area Between The Curve Problem No 1 - Applications Of Definite Integration - Diploma Maths II; Author: Ekeeda;https://www.youtube.com/watch?v=q3ZU0GnGaxA;License: Standard YouTube License, CC-BY