A 6.5 kg particle starts from rest at x = 0 & moves under the influence of a single force
Fx = 6.1 + 7.6x - 1.1x^2, and Fx is in Newtons and x is in meters. Find the work done by ths force on the particle as the particle moves from x = 0 m to x = 2.2m. Answer in units of J???
2006-10-02 08:21:53 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Physics
The work done on an object by a force is given by the integral of F (dot) ds, F is the force vector, and ds is the infinitesimal displacement vector along the path of motion of the object. The limits of integration extend from the starting to the ending location of the object.
The mass of the object is irrelevant if you are given the force
In this case, the direction of the applied force is not provided (i.e., you haven't given F in the form of a vector), so I'll assume that the force is directed along the x-axis. In that case, the motion will also be restricted to the x-axis.
The work done is then:
W = Integral from 0 to 2.2 m of {(6.1 + 7.6s - 1.1s^2) ds}
W = 6.1*2.2 + 3.8*(2.2)^2 - (1.1/3)*(2.2)^3 - 0 J
W = 27.908 J
2006-10-02 08:39:19 · answer #1 · answered by hfshaw 7 · 0⤊ 0⤋
Work = force * distance
Integral from 0 to 2.2 of Fx dx
Doug
2006-10-02 08:32:30 · answer #2 · answered by doug_donaghue 7 · 0⤊ 1⤋
fairly, in #a million, for each 2 ft in genuine life, on the blueprint there is one inch. so which you divide the 8 by potential of two, 4 inches, and 12 by potential of two, 6 inches. For #2, comparable element, yet divide 8 and 12 by potential of three. #3 divide by potential of four and #4 divide by potential of two.5 i think of.
2016-10-15 10:53:16 · answer #3 · answered by ? 4 · 0⤊ 0⤋
179.974 J
Integrate the equation from 0 to 2.2
multiply the integration by the mass and....voila
2006-10-02 08:35:19 · answer #4 · answered by Anonymous · 0⤊ 1⤋