Question

A. 740-kg elevator accelerated upward at 1.1m/s^2 ,pulled by a cable of negligible mass. Find the tension force in the cable.

Answer 1

Given:-

• Mass of elevator ,m = 740 Kg

• Acceleration ,a = 1.1 m/s²

To Find:-

• Tension Force in the Cable ,T

Solution:-

We have to calculate the tension force in the cable .

As we know that

• F = ma

Where,

F is the Force

m is the mass

a is the acceleration

and Also

• T -mg = ma

→ T = mg + ma

→ T = m(g+a)

where,

T is the tension

m is the mass

g is the acceleration due to gravity = 9.8m/s²

a is the acceleration.

Substitute the value we get

→ T = 740 (9.8+1.1)

→ T = 740 (10.9)

→ T = 740 × 10.9

→ T = 8066 N

Therefore, the tension force in the cable is 8066 Newton .

Answer 2

Answer:

Given :-

• Mass of elevator (M) = 740 kg
• Acceleration (A) = 1.1m²
• Acceleration due to gravity (G) = 9.8 m/s²

To Find :-

Tension force

Solution :-

As we know that

$\huge \bf \: T = m(g + a)$

Here,

T = Tension force

M = Mass

G = Acceleration due to gravity

A = Acceleration

Putting values

$\sf \: T = 740(9.8 + 1.1)$

$\sf \: T = 740 \times 10.9$

$\sf \: T = 8066 \: N$

Therefore,

Tension force of elevator = 8066 Newton.