An 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.
The pull of cable is 5.24x10^3N
The force exerted by the man =75N
Is it correct?
first, find the acceleration.
Vf^2=Vi^2+2ad
4=0+2a*3
a=2/3 m/s^2
then
force= m(g+a)=500(9.8+2/3) =
To find the pull in the cables during the elevator's accelerated motion, we need to use Newton's second law of motion.
1. Find the net force acting on the elevator during the accelerated motion.
- The net force (F_net) can be calculated using the formula:
F_net = m * a
where m is the mass of the elevator, and a is the acceleration.
- Given that the mass of the elevator (m) is 500 kg and the elevator acquires a velocity of 2 m/s after traveling a distance of 3 m, we can find the acceleration using the kinematic equation:
v^2 = u^2 + 2as
where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance traveled.
- Plugging in the given values:
2^2 = 0 + 2a * 3
4 = 6a
a = 4/6 = 0.67 m/s^2
- Calculating the net force:
F_net = m * a
F_net = 500 kg * 0.67 m/s^2
F_net = 335 N
Therefore, the pull in the cables during the elevator's accelerated motion is 335 N.
2. To calculate the force exerted by the man on the elevator floor during the deceleration, we can again use Newton's second law of motion.
- The force exerted by the man (F_man) on the elevator floor can be calculated using the formula:
F_man = m * a
where m is the mass of the man, and a is the acceleration due to deceleration.
- Given that the mass of the man (m) is 75 kg and the elevator comes to rest in 2 seconds, we can find the acceleration using the formula:
a = (v - u) / t
where v is the final velocity, u is the initial velocity, and t is the time taken.
- Plugging in the given values:
a = (0 - 2 m/s) / 2 s
a = -2 m/s^2
- Calculating the force exerted by the man:
F_man = m * a
F_man = 75 kg * -2 m/s^2
F_man = -150 N (negative sign indicates the direction of force opposite to motion)
Therefore, the force exerted by the man on the elevator floor during the deceleration is 150 N (opposite to the motion).
To calculate the pull in the cables during the accelerated motion, we need to consider the forces acting on the elevator.
Let's break down the problem step by step:
Step 1: Calculate the acceleration during the upward motion.
We know that the elevator's initial velocity is 0 m/s because it starts from rest. The final velocity is 2 m/s, and the displacement is 3 m. We can use the following kinematic equation to find the acceleration:
v^2 = u^2 + 2as
where:
v = final velocity = 2 m/s
u = initial velocity = 0 m/s
a = acceleration
s = displacement = 3 m
Rearranging the equation, we get:
a = (v^2 - u^2) / (2s)
= (2^2 - 0^2) / (2 * 3)
= 4 / 6
= 2/3 m/s^2
So, the acceleration during the upward motion is 2/3 m/s^2.
Step 2: Calculate the net force during the accelerated motion.
The net force is given by Newton's second law: F = ma
The mass of the elevator is given as 500 kg, so the net force is:
F = m * a
= 500 kg * 2/3 m/s^2
= 1000/3 = 333.33 N
Therefore, the pull in the cables during the accelerated motion is 333.33 N.
Now, let's move on to the second part of the question regarding the force exerted by the man on the floor during the deceleration.
Step 3: Calculate the deceleration during the downward motion.
The elevator is moving with constant deceleration, so the deceleration is equal to the opposite of the acceleration during the upward motion.
Deceleration = -2/3 m/s^2 (negative because it is in the opposite direction)
Step 4: Calculate the force exerted by the man.
In this scenario, the man's weight is the force exerted by him on the floor. Weight is given by the formula:
Weight = mass * acceleration due to gravity
The mass of the man is given as 75 kg, and acceleration due to gravity is approximately 9.8 m/s^2.
Weight = 75 kg * 9.8 m/s^2
= 735 N
Therefore, the force exerted by the man on the floor of the elevator is 735 N.