Thank You A sign in a 5.00 * 102 kg elevator states that the maximum occupancy i

Question

Thank You

A sign in a 5.00 * 102 kg elevator states that the maximum occupancy is 20 persons. If the cable supporting the elevator can tolerate a maximum force of 3.0 * 10A N, what is the greatest acceleration that the elevator's motor can produce without snapping the cable? (Assume the average person has 75 kg of mass) A racecar has a mass of 710 kg and it can travel 40.0 meters in just 3.0 seconds, starting from rest. If the car is experiencing uniform acceleration for the duration of movement, determine the net force acting on the car.

Explanation / Answer

(3)

Equating the vertical forces on the Elevator

T - (M + n * m) * g = (M + n * m) * a

=> T = (M + n * m) * (g + a)

=> g + a = T / (M + n * m)

=> a = [T / (M + n * m)] - g

Where n is the number of the persons

m is the mass of each person

M is the mass of the lift

a is the greatest acceleration due to gravity possible

T is the maximum force the cable can withstand

Substituting the values

a = [3 * 10^4 / (2 * 10^3)] - 9.81

a = 5.19 m / s^2

The maximum acceleration the cable can tolerate is 5.19 m / s^2

(4)

Given initial velocity u = 0 m/s

distance travelled x = 40 meters

time taken = 3 seconds

From x = u * t + 0.5 * a * t^2

Substituting

40 = 0 + 0.5 * a * 9

=> a = 8.89 m/s^2

Net force acting on the car = m * a

=> F = 710 * 8.89 = 6311.11 N

Net force acting on the car is 6311.11 N

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