Answer:
Approximately [tex]6.04\; {\rm m\cdot s^{-2}}[/tex], assuming that [tex]g = 9.81\; {\rm N \cdot kg^{-1}}[/tex] and that the mass of the bag is negligible.
Explanation:
There are two forces on the paper bag:
It is given that the tension in the handle [tex]F(\text{tension})[/tex] should not exceed [tex]298\; {\rm N}[/tex].
Let [tex]g[/tex] denote the gravitational field strength. The mass of the bag and its contents is [tex]m = 18.8\; {\rm kg}[/tex]. Their weight will be [tex]F(\text{weight})= m\, g[/tex].
SInce [tex]F(\text{tension})[/tex] and [tex]F(\text{weight})[/tex] are in opposite directions, the resultant force on the bag will be:
[tex]F(\text{net}) = F(\text{tension}) - F(\text{weight})[/tex].
Divide the net force by mass to find the acceleration [tex]a[/tex] of the bag:
[tex]\begin{aligned}a &= \frac{F(\text{net})}{m} \\ &= \frac{F(\text{tension}) - F(\text{weight})}{m} \\ &= \frac{F(\text{tension}) - m\, g}{m} \\ &= \frac{F(\text{tension})}{m} - g\end{aligned}[/tex].
Since [tex]F(\text{tension}) \le 298\; {\rm N}[/tex]:
[tex]\begin{aligned} a &= \frac{F(\text{tension})}{m} - g \\ &\le \frac{298\; {\rm N}}{18.8\; {\rm kg}} - 9.81\; {\rm N \cdot kg^{-1}} \\ &\approx 6.04\; {\rm N \cdot kg^{-1}} \\ &=6.04\; {\rm m\cdot s^{-2}}\end{aligned}[/tex].
Note the unit conversion: [tex]1\; {\rm N \cdot kg^{-1}} = 1\; {\rm (kg \cdot m\cdot s^{-2})\, kg^{-1}} = 1\; {\rm m\cdot s^{-2}}[/tex].
Answer:
Acceleration = 6.05 m/s²
Explanation:
Fmax = 298 N
m = 18.8 kg
g = 9.8 m/s²
___________
a - ?
According to Newton's 2nd law:
Fmax = m·(g + a)
g + a = Fmax / m
Acceleration:
a = Fmax / m - g
a = 298 / 18.8 - 9.8 ≈ 6.05 m/s²
