Question

How much tension must a cable withstand if it is used to accelerate a 1800- kg car vertically upward at 0.60 m/s2

Answer 1

Approximately
`1.9* 10^(4)\; {\rm N}` (rounded up) assuming that
`g = (-9.81)\; {\rm N \cdot kg^(-1)}` (downwards.)

Step-by-step explanation:

The car is accelerating at a constant
`a = 0.60\; {\rm m\cdot s^(-2)}` (positive since the car is accelerating upwards.) Hence, the net force on this car will be:

`\begin{aligned}F_{\text{net}} &= m\, a \\ &= 1800\; {\rm kg} * 0.60\; {\rm m\cdot s^(-2)} \\ &= 1080\; {\rm N}\end{aligned}`.

Note that since net force points in the same direction as acceleration (upwards,)
`F_{\text{net}}` is also positive.

The main forces on this car are:

• Weight (downward).
• Tension from the cable (upward):
  `F(\text{tension})`.

With a mass of
`m = 1800\; {\rm kg}`, the weight on this car will be
`m\, g = 1800\; {\rm kg} * (-9.81)\; {\rm N \cdot kg} = (-17658)\; {\rm N}` (negative since weight points downwards to the ground.)

The net force on this car is the sum of the external forces:

`F_{\text{net}} = F(\text{tension}) + m\, g`.

It is shown that
`F_{\text{net}} = 1080\; {\rm N}` while
`m\, g = (-17658)\; {\rm N}`. Substitute both values into the equation and solve for
`F(\text{tension})`:

`1080\; {\rm N} = F(\text{tension}) + (-17658)\; {\rm N}`.

`\begin{aligned}F(\text{tension}) &= 1080\; {\rm N} + 17658\; {\rm N} \approx 1.9* 10^(4)\; {\rm N} \end{aligned}` (rounded up.)