1. An elevator weighing 15 kN starts from rest and acquires an upward velocity of 3 m/sec. in a distance of 6 m. If the acceleration is constant, what is the tension in the cable?

2. A man weighing 70 kg is in an elevator moving upward with an acceleration of 2.5 m/sec2.

A. What pressure does he exert on the floor of the elevator?
B. What will be the pressure if the elevator is descending down with the same acceleration?

16.15 kN

61.65234

To solve these problems, we will need to use the equations of motion and the concept of Newton's second law.

1. To find the tension in the cable, we can use the equation:
\(F = ma\), where F is the net force acting on the elevator, m is the mass of the elevator, and a is the acceleration of the elevator.

First, we need to find the mass of the elevator using the equation:
\(F = mg\), where g is the acceleration due to gravity (approximately 9.8 m/s²).
Rearranging the equation, we get:
\(m = \frac{F}{g}\)
Substituting the given value of the weight, F = 15 kN = 15,000 N, we have:
\(m = \frac{15,000 N}{9.8 m/s²} \approx 1531.6 kg\)

Now we can find the tension in the cable using the equation:
\(F = ma\)
Substituting the values of m = 1531.6 kg and a = 3 m/s², we have:
\(F = (1531.6 kg)(3 m/s²) \approx 4594.8 N\)

Therefore, the tension in the cable is approximately 4594.8 N.

2. A. To find the pressure exerted by the man on the floor of the elevator, we can use the equation:
\(F = ma\), where F is the force exerted by the man, m is the mass of the man, and a is the acceleration of the elevator.

First, we need to find the force exerted by the man using the equation:
\(F = mg\), where g is the acceleration due to gravity (approximately 9.8 m/s²).
Substituting the given value of the man's weight, m = 70 kg, we have:
\(F = (70 kg)(9.8 m/s²) = 686 N\)

Now we can find the pressure exerted by the man using the equation:
\(P = \frac{F}{A}\), where P is the pressure, F is the force, and A is the area.

Since pressure is defined as force per unit area, we need to consider the area over which the pressure is exerted. In this case, it is the contact area between the man's feet and the floor of the elevator.

Without the specific area value, we cannot calculate the exact pressure exerted.

2. B. If the elevator is descending with the same acceleration as before, the pressure exerted by the man on the floor would be different. The equation to calculate the pressure remains the same:
\(P = \frac{F}{A}\)

However, as the acceleration is now downward, we need to consider the effective weight of the man. The net force acting on the man is the difference between his weight and the force due to acceleration.
\(F = mg - ma\)

Substituting the given values of m = 70 kg and a = 2.5 m/s², we have:
\(F = (70 kg)(9.8 m/s²) - (70 kg)(2.5 m/s²) = 637 N\)

Now we can calculate the pressure exerted by the man using the equation:
\(P = \frac{F}{A}\)

Without the specific area value, we cannot calculate the exact pressure exerted.

An elevator weighing 980N attains an upward velocity of 3m/s in 3s following a uniform acceleration. The tension in the cable that supports the elevator is