Question

An elevator and its load have a total mass of 800 kg. The elevator is originally moving downward at 10 m/s, it slows down to stop with constant acceleration at a distance of 25 m. Find the tension t in the supporting cable while the elevator is being brought to rest.

Answer 1

Answer:

9600 Newtons [If this answer helps you then please show your gratitude by marking this answer as the brainliest answer.]

Explanation:

u(Initial Velocity)=10 m/s

v(Final Velocity)=0 m/s

s=25 m

t=?

Answer 2

Given :-

Mass of Elevator along its load = m = 800 Kg

Initial Velocity = u = 10 ms-¹

Heights travelled = h = 25 m

Acceleration due to gravity = g = 10 ms-².

Using third equation of Motion to find acceleration of the body.

v² = u² - 2ah

u² = 2ah

a = u²/2h

a = 100/50

a = 2 ms-².

${R}_{u}$ = m(g+a)

Let the tension R is acting in upward direction.

${R}_{u}$ = 800(10+2)

${R}_{u}$ = 9600 N