Answer:
Approximately [tex]1.9\times 10^{4}\; {\rm N}[/tex] (rounded up) assuming that [tex]g = (-9.81)\; {\rm N \cdot kg^{-1}}[/tex] (downwards.)
Explanation:
The car is accelerating at a constant [tex]a = 0.60\; {\rm m\cdot s^{-2}}[/tex] (positive since the car is accelerating upwards.) Hence, the net force on this car will be:
[tex]\begin{aligned}F_{\text{net}} &= m\, a \\ &= 1800\; {\rm kg} \times 0.60\; {\rm m\cdot s^{-2}} \\ &= 1080\; {\rm N}\end{aligned}[/tex].
Note that since net force points in the same direction as acceleration (upwards,) [tex]F_{\text{net}}[/tex] is also positive.
The main forces on this car are:
With a mass of [tex]m = 1800\; {\rm kg}[/tex], the weight on this car will be [tex]m\, g = 1800\; {\rm kg} \times (-9.81)\; {\rm N \cdot kg} = (-17658)\; {\rm N}[/tex] (negative since weight points downwards to the ground.)
The net force on this car is the sum of the external forces:
[tex]F_{\text{net}} = F(\text{tension}) + m\, g[/tex].
It is shown that [tex]F_{\text{net}} = 1080\; {\rm N}[/tex] while [tex]m\, g = (-17658)\; {\rm N}[/tex]. Substitute both values into the equation and solve for [tex]F(\text{tension})[/tex]:
[tex]1080\; {\rm N} = F(\text{tension}) + (-17658)\; {\rm N}[/tex].
[tex]\begin{aligned}F(\text{tension}) &= 1080\; {\rm N} + 17658\; {\rm N} \approx 1.9\times 10^{4}\; {\rm N} \end{aligned}[/tex] (rounded up.)
