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Here are examples of how to use Newton's second law for a quck when one or two forces act on it. Parts (a), (b), and ( c ) of Fig. 5-3 show three situations in which one or two forces act on a puck that moves over frictionless ice along an x axis, in one-dimensional motion. The puck's mass is m = 0.20 kg. Forces vec(F)_(1) and vec(F)_(2) are directed along the axis and have magnitudes F_(1)=4.0 N and F_(2)=2.0 N. Force vec(F)_(3) is directed at angle theta=30^(@) and has magnitude F_(3)=1.0 N. In each situation, what is the acceleration of the puck ? |
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Answer» Solution :In each situation we can relate the ACCELERATION `vec(a)` to the net force `vec(F)_("net")` acting on the puck with Newton.s second law, `vec(F)_("net")=m vec(a)`. However, because the motion is along only the x axis, we can simplify each situation by writing the second law for x components only: `F_("net.x")=ma_(x).""(5-4)` The free-body diagrams for the three situations are also given in FIG. 5-3, with thee puck represented by a dot. Calculations : Situation A: Foor Fig. 5-3b, where only one horizontal force acts, Eq. 5-4 gives us `F_(1)=ma_(x)`, which, with given data, yields `a_(x)=(F_(1))/(m)=(4.0N)/(0.20kg)=20" m"//"s"^(2)`. The positive answer indicates that the acceleration is in the positive direction of the x axis. Situation B: In Fig. 5-3d, two horizontal forces act on the puck, `vec(F)_(1)` in the positive direction of x and `vec(F)_(2)` in the NEGATIVE direction. Now Eq. 5-4 gives us `F_(1)-F_(2)=ma_(x)`, which, with given data, yields `a_(x)=(F_(1)-F_(2))/(m)=(4.0N-2.0N)/(0.20kg)=10" m"//"s"^(2)`. Thus, the net force accelerates the puck in the positive direction of the x axis. Situation C: In Fig. 5-3f, force `vec(F)_(3)` is not directed along the direction oof the puck.s acceleration, only x component `F_(3,x)` is. (Force `vec(F)_(3)` is two-dimensional but the motion is only one-dimensional.) Thus, we write Eq. 5-4 as  FIGURE 5-3In three situations, forces act on a puck that moves along an x axis. Free-body diagrams are also shown. `F_(3,x)-F_(2)=ma_(x).""(5-5)` From the figure, we see that `F_(3,x)=F_(3)cos theta`. Solving for the acceleration and substituting for `F_(3,x)` yield `a_(x)=(F_(3,x)-F_(2))/(m)=(F_(3)cos theta-F_(2))/(m)` `=((1.0N)(cos 30^(@))-2.0N)/(0.20kg)=-5.7" m"//"s"^(2).` Thus, the net force accelerates the puck in the negative direction of the x axis. |
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