How much tension must a rope withstand if it is used to accelerate a 1210-kg car horizontally along a frictionless surface at$a\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{20}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$.

A car having some mass is being accelerated with the help of a rope in the horizontal direction.

Therefore, the force applied to the car is equal to the tension in the rope. The relation can be obtained with the help of Newton's second law.

The given data can be listed below as:

• The mass of the car is $m=1210\mathrm{kg}$.
• The acceleration of the car is $a=1.20\mathrm{m}/{\mathrm{s}}^2$.
• The surface is frictionless; so the friction value is zero.

From Newton’s second law, the force applied on the car with the help of a rope to accelerate it can be expressed as:

$F=ma$

Here, m is the mass of the car, and a is the acceleration of the car.

Substitute the values as 1210 kg for m, and $1.20\mathrm{m}/{\mathrm{s}}^2$for a in the above equation.

$\begin{array}{c}F=1210\mathrm{kg}\times 1.20\mathrm{m}/{\mathrm{s}}^2\left(\frac{1\mathrm{N}}{1\mathrm{kg}·\mathrm{m}/{\mathrm{s}}^2}\right)\\ =1452\mathrm{N}\end{array}$

The force applied to the car is equal to the tension force that acts in the rope.

$\begin{array}{c}F=T\\ =1452\mathrm{N}\end{array}$

Thus, the tension that a rope must withstand is $1452\mathrm{N}$.