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In Fig. 5-29, a block of mass m = 5.00 ...

In Fig. 5-29, a block of mass m = 5.00 kg is pulled along a horizonttal frictionless floor by a cord that exerts a force of magnitude F = 12.0 at an angle `theta=25.0^(@)`.

Figure 5-29 A block is pulled along a horizontal floor.
(a) What is the magnitude of the block's acceleration?

Text Solution

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The free-body diagram (not to scale) for the block is shown in Fig. 5-30. `vec(F)_(N)` is the normal force exerted by the floor and `m vec(g)` is the force of gravity.
Calculation: The x component of Newton.s second law is `F cos theta=ma`, where m is the mass of the block and a is the x component of its acceleration. We obtain
`a=(F cos theta)/(m)=((12.0" N")cos 25.0^(@))/(5.00" kg")=2.18"m"//"s"^(2)`.
This is its acceleration provided it remains in contact with the floor. Assuming it does, we find the value of `F_(N)` (and if `F_(N)` is positive, then the assumption is true but if `F_(N)` is negative then the block leaves the floor). The y component of Newton.s second law becomes
`F_(N)+F sin theta-mg=0`,
so
`F_(N)=mg-F sin theta`
`=(5.00" kg")(9.80" m"//"s"^(2))-(12.0" N")sin 25.0^(@)`
`=43.9" N"`.
Hence, the block remains on the floor and its acceleration is `a=2.18" m"//"s"^(2)`.
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