n elevator with a man inside, of gross mass 500KG, starts moving upwards with a costant acceleration and acquires a velocity os 2m/s after travelling a distance os 3m.Find the pull in the cables during the accelerated motion.
If the elevator, before stopping, moves with a constant deceleration from a constant velocity of 2m/s and comes to rest in 2 second, calculate the force exerted by the man of 75kg mass on the floor on the elevator.
To find the pull in the cables during the accelerated motion of the elevator, we need to calculate the net force acting on the elevator.
Step 1: Calculate the acceleration of the elevator.
Using the equation of motion: v^2 = u^2 + 2as, where:
- v = final velocity = 2 m/s
- u = initial velocity = 0 m/s (stationary)
- s = distance traveled = 3 m
Rearranging the equation, we get: 2^2 = 0^2 + 2a(3)
4 = 6a
a = 4/6
a = 0.67 m/s^2
Step 2: Calculate the net force.
The net force acting on an object is given by Newton's second law of motion (F = ma), where:
- F = net force
- m = mass of the object
F = m * a
F = 500 kg * 0.67 m/s^2
F = 335 N
So, the pull in the cables during the accelerated motion of the elevator is 335 Newtons.
Now, let's move on to the second part of the question.
To calculate the force exerted by the man on the floor of the elevator during the deceleration, we need to find the net force acting on the man.
Step 1: Calculate the acceleration during deceleration.
Using the equation of motion: v = u + at, where:
- v = final velocity = 0 m/s (stopped)
- u = initial velocity = 2 m/s
- t = time taken = 2 s
Rearranging the equation, we get: 0 = 2 + a(2)
-2 = 2a
a = -1 m/s^2
(Note the negative sign indicates deceleration)
Step 2: Calculate the net force.
F = m * a
F = 75 kg * (-1 m/s^2)
F = -75 N
(Note the negative sign indicates the direction of force opposing the motion)
Therefore, the force exerted by the man on the floor of the elevator is -75 Newtons (opposite to the direction of motion).