✔ 最佳答案
Work done = Fs = Fs cosθ
Yet since WD is a kind of energy, it can also be equalised to KE or KE and PE sometimes.
***NOTE that if there is NO change in mechanical energy, there is NO WD.
It depends on the angle made between the direction of movement and the y-component applied force (to move the object), as I remember.
There are 3 cases (4, if being more strict):
Let the applied force be 5 N and the displacement of the object (no matter in what direction) be 2 m.
1) Moving laterad (to the right or left: → / ←)
(i) If the force is PARALLEL to the direction of movement,
WD = Fs cos0
= Fs(1)
= Fs
= 5(2) = 10 J
(ii) If the force is NOT parallel to the direction of movement but, say, is making an angle of 60 deg. with it,
WD = Fs cos60deg.
= 1/2(Fs)
= (1/2)(5)(2) = 5 J
On the other hand, the object moves sidewards, so it had gained some KE from the WD, i.e. WD can be expressed as GAIN in KE as well.
(***Only the y-component of the force contributes to the change in KE, i.e. change in mechanical energy)
2) Moving vertically (upwards or downwards: ↑ / ↓)
(i)If the force applied is PERPENDICULAR to the direction of movement,
WD = Fs cos90deg.
= Fs(0)
= NO WD
Whenever the applied force is perpedicular to the direction of movement, i.e. making an angle of 90deg., there will be no WD.
Although the object has gained both KE and PE, the changes in energy are NOT contributed by the y-component of the force, thus we say that there is no work done.
3) Moving diagonally (BOTH vertically and horizontally: ↘ /↙ /↗ /↖)
(i) If the force makes an angle of 60deg. with the direction of movement and the object moves 2 m to the right upwards↗,
WD = PE + KE
= mgh + (1/2)mv^2
Since the applied force is not perpendicular nor parallel to the direction of movement, and the object has gained both kinetic and potential energy under the act of the y-component of the force.
Your question should be referring to the 3)(i) case.
參考: moi, hope it helps.=]] Please kindly point out any mistakes spotted.