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leannhb3162
03.03.2020 •
Physics
A constant force of 12 N in the +x direction acts on a 4 kg block as it moves from the origin to the point (6 − 8) m. How much work is done on the block by this force during this displacement?
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Ответ:
72 Nm²
Explanation:
work = F . d
work = (12i ,0j) . (6i ,-8j )
work = 72 J
The 12N force is a vector acting in the direction of the positive x axis, so its vector notation is 12i + 0j.
It's not exactly clear what the point's location is but I interpret it to be x= m*6, y= -m*8 (I hope the "m" wasn't a typo). The direction vector from the origin is then m*6i -m*8j
When a force vector F acts in a nonparallel direction d, the work is given by:
W = |F|*|d|*cos(theta) where theta is the angle between the vectors.
alternatively you can use the dot product of the two vectors to get:
W = (12N)*(m*6) + (0N)*(-m*8) = 72 Nm²
Ответ:
100j
Explanation: