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Sagot :
Answer:
Approximately [tex]33\; {\rm m\cdot s^{-2}}[/tex].
Approximately [tex]55\; {\rm m\cdot s^{-2}}[/tex].
(Assume that the table is level, and that the [tex]60\; {\rm N}[/tex] force is horizontal.)
Explanation:
Consider all three blocks as one object of mass [tex]m = (1 + 4 + 6)\; {\rm kg} = 11\; {\rm kg}[/tex]. Among all the forces that are in action, the only unbalanced external force on this [tex]m = 11\; {\rm kg}[/tex] object will be the [tex]60\; {\rm N}[/tex] force. Hence, the resultant force of this combined object of mass [tex]m = 11\; {\rm kg}[/tex] will be [tex]F_{\text{net}} = 60\; {\rm N}[/tex].
Acceleration [tex]a[/tex] of this combined object will be:
[tex]\begin{aligned}a &= \frac{F_{\text{net}}}{m} \\ &= \frac{60\; {\rm N}}{11\; {\rm kg}} \\ &= \frac{60}{11}\; {\rm m\cdot s^{-2}}\end{aligned}[/tex].
Since the three crate blocks are moving together, each will have the same acceleration, [tex]a = (60/11)\; {\rm m\cdot s^{-2}}[/tex].
Resultant force on each of the crate blocks will be:
- [tex]1\; {\rm kg}[/tex] crate: [tex]F_{\text{net}} = m\, a = (1\; {\rm kg})\, (60/11\; {\rm m\cdot s^{-2}}) = (60/11)\; {\rm N}[/tex].
- [tex]4\; {\rm kg}[/tex] crate: [tex]F_{\text{net}} = m\, a = (4\; {\rm kg})\, (60/11\; {\rm m\cdot s^{-2}}) = (240/11)\; {\rm N}[/tex].
- [tex]6\; {\rm kg}[/tex] crate: [tex]F_{\text{net}} = m\, a = (6\; {\rm kg})\, (60/11\; {\rm m\cdot s^{-2}}) = (360/11)\; {\rm N}[/tex].
Assume that the [tex]60\; {\rm N}[/tex] external force on the [tex]1\; {\rm kg}[/tex] block points to the right.
When the crates are considered individually, external forces on the [tex]1\; {\rm kg}[/tex] crate will include:
- the [tex]60\; {\rm N}[/tex] external force to the right, and
- a normal force the [tex]4\; {\rm kg}[/tex] block exerts on the [tex]1\; {\rm kg}[/tex] block (to the left.) Assume that this force is of magnitude [tex]x\; {\rm N}[/tex].
- (In the vertical direction, the weight of this block and the upward normal force from the table are balanced.)
Since these two forces are in opposite directions, the resultant force on this [tex]1\; {\rm kg}[/tex] block will be [tex](60\; {\rm N} - x\; {\rm N})[/tex]. However, since the actual resultant force on this block (calculated from acceleration) is [tex](60 / 11)\; {\rm N}[/tex]:
[tex]\displaystyle 60\; {\rm N} - x\; {\rm N} = \frac{60}{11}\; {\rm N}[/tex].
Therefore, the force that the [tex]4\; {\rm kg}[/tex] block exerts on the [tex]1\; {\rm kg}[/tex] block will be
[tex]\displaystyle 60\; {\rm N} - \frac{60}{11}\; {\rm N} = \frac{600}{11}\; {\rm N} \approx 55\; {\rm N}[/tex].
When considered individually, the only unbalanced external force on the [tex]m = 6\; {\rm kg}[/tex] block is the normal force from the [tex]4\; {\rm kg}[/tex] block. Hence, this force will be equal to the resultant force on the [tex]m = 6\; {\rm kg}[/tex] block, [tex](360 / 11)\; {\rm N} \approx 33\; {\rm N}[/tex].
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