Among the safety features on elevator cages are spring-loaded brake pads which grip the guide rail if the elevator cable should break. Suppose that an elevator cage of 2000 kg has two such pads, arranged to press against opposite sides of the guide rail, each with a force of 8.00 104 N. The friction coefficient for the brake pads sliding on the guide rail is 0.15. Assume that the elevator cage is falling freely with an initial speed of 10.5 m/s when the brake pads come into action. How long will the elevator cage take to stop (in seconds)? How far will it travel (in m)? How much energy is dissipated by friction? (in J)
To find the answers to these questions, we need to analyze the forces acting on the elevator cage and use the principles of Newtonian mechanics. The key forces involved are the force of gravity and the frictional force acting on the brake pads.
1. How long will the elevator cage take to stop (in seconds)?
To determine the time it takes for the elevator cage to stop, we can use the equation for acceleration:
a = (ΣF) / m
where a is the acceleration, ΣF is the sum of the forces acting on the cage, and m is the mass of the cage.
The sum of the forces acting on the cage includes the gravitational force and the frictional force:
ΣF = mg - μN
where g is the acceleration due to gravity, μ is the friction coefficient, and N is the normal force.
The normal force N is equal to the force applied by each brake pad:
N = F
substituting the values given:
N = 8.00 * 10^4 N
Now, we can calculate the frictional force:
f = μN
substituting μ = 0.15:
f = 0.15 * (8.00 * 10^4 N)
Next, we can subtract the frictional force from the gravitational force to find the net force:
ΣF = mg - f
substituting the mass given (m = 2000 kg) and g (approximately 9.8 m/s^2):
ΣF = (2000 kg * 9.8 m/s^2) - f
Now, we can calculate the acceleration (a) using the first equation:
a = ΣF / m
2. How far will it travel (in m)?
To determine the distance traveled by the elevator cage before it stops, we need to calculate the deceleration. Since the cage starts with an initial speed and then comes to rest, we can use the equation:
v^2 = u^2 + 2as
where v is the final velocity (0 m/s when the elevator stops since it comes to rest), u is the initial velocity, a is the acceleration, and s is the distance traveled.
Solving for s:
s = (v^2 - u^2) / (2a)
substituting the values:
s = (0^2 - (10.5 m/s)^2) / (2 * a)
3. How much energy is dissipated by friction? (in J)
To calculate the energy dissipated by friction, we need to find the work done by the frictional force. The work done by a force is given by the equation:
W = F * d
where W is the work done, F is the force, and d is the distance.
In this case, the force is the frictional force (f) and the distance is the distance traveled (s) before stopping.
W = f * s
substituting the calculated values for f and s, we can find the energy dissipated by friction.
vf=vi-2at but a= force/mass= 2*8e4/2000=80 m/s^2
so figure time t, if vf=0
distance: avg speed= vi/2, so distance= avgspeed*time