Question

An 888.0 kg elevator is moving downward with a velocity of 0.800 m/s. It decelerates uniformly and comes to a stop in a distance of 0.2667 m. What is the tension on the cable while it decelerates

Answer 1

Answer:

The value of tension on the cable T = 1065.6 N

Explanation:

Mass = 888 kg

Initial velocity ( u )= 0.8 $\frac{m}{sec}$

Final velocity ( V ) = 0

Distance traveled before come to rest = 0.2667 m

Now use third law of motion $V^{2}$ = $u^{2}$ - 2 a s

Put all the values in above formula we get,

⇒ 0 = $0.8^{2}$ - 2 × a ×0.2667

⇒ a = 1.2 $\frac{m}{sec^{2} }$

This is the deceleration of the box.

Tension in the cable is given by T = F = m × a

Put all the values in above formula we get,

T = 888 × 1.2

T = 1065.6 N

This is the value of tension on the cable.