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Here we calculate on instantaneous work - that is, the rate at which work is being done at any given instant rather than averaged over a time interval. Figure 8-34 shows constant forces vec(F)_(1) and vec(F)_(2) acting onabox as the box slides rightward across a frictionless floor. Force vec(F)_(1) is horizontal , with magnitude 2.0 N, force vec(F)_(2) is angled upward by 60^(@) to the floor and has magnitude 4.0 N. The speed v of the box at a certain instant is 3.0 m/s. What is the power due to each force acting on the box at that instant, and what is the net power? Is the net power changing at that instant? |
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Answer» Solution :KEY IDEAS We want an instantaneous power, not an average power over a time period. Also, we KNOW the box.s velocity (rather than the work done on it). Calculation: We use Eq. 8-87 for each force, For force `vec(F)_(1)` at angle `phi_(1)=180^(@)` to velocity `vec(v)` , we have `P_(1) = F_(1) v cos phi_(1) = (2.0N)(3.0 m//s) cos 180^(@)` ` = -6.0` W. This negative result tells us that force `vec(F)_(1)` is TRANSFERRING energy from the box at the rate of 6.0 J/s. For force `vec(F)_(2)`, at angle `phi_(2)=60^(@)` to velocity `vec(v)`, we have `P_(2)= F_(2) v cos phi_(2)=(4.0 N) (3.0 m//s) cos 60^(@)` `= 6.0 W`.  Figure 8-34 Two forces `vec(F)_(1)` and `vec(F)_(2)` act on a box that slides reightward across a frictionless floor. The velocity of the box is `vec(v)`. This positive result tells us that force `vec(F)_(2)` is transferring energy to the box at the rate of 6.0 J/s. The net power is the sum of theindividual powers ( complete with their algebraic SIGNS): ` P_("net")=P_(1)+P_(2)` ` = -60 W + 6.0 W = 0`, Which tells us that the net rate of transfer of energy to or fromthe box is zero. Thus, the kinetic energy `(K=1//2 mv^(2))` of the box is not changing, and so the speed of the box will remain at 3.0 m//s. With neither the forces `vec(F)_(1)` and `vec(F)_(2)` nor HTE velocity `vec(v)` changing , We see from Eq. 8-88 that `P_(1)` and `P_(2)` are constant and thus So is `P_("net")`. |
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