Question

How much tension must a rope withstand if it is used to accelerate a 1210- kg car horizontally along a frictionless surface at 1.20 m/s squared?

Answer 1

Newton's second law states that the force F applied to the car is the product between the mass of the car m and its acceleration a:
$F=ma$
But the force applied to the car is the tension of the rope, T, so we have:
$T=ma$

And so, using the data of the problem we calculate the value of the tension:
$T=ma=(1210 kg)(1.20 m/s^2)=1452 N$

Answer 2

The rope can withstand the tension of 1452 N.

Given data:

The mass of car is, m = 1210 kg.

The acceleration of car is, $a = 1.20 \;\rm m/s^{2}$.

Since tension force is acting on the rope which provides an external force to accelerate the car. Then, the tension acting on the rope is equivalent to,

T = F ...................................................(1)

Here, F is the external force causing the acceleration of car. And its value is given as,

F = ma .......................................................(2)

Substitute the value of equation 2 in 1 as,

T= F

T = ma

$T = 1210 \times 1.20 \\T =1452 \;\rm N$

Thus, the tension force acting on the rope is 1452 N.

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