A descending elevator of mass 670 kg is uniformly decelerated to rest over a distance of6 m by a cable in which the tension is 8683 N.The acceleration due to gravity is 9.8 m/s2.Calculate the speed vi of the elevator at the beginning of the 6 m descent. Answer in units of m/s.
The deceleration rate a, stopping distance X and the initial velocity Vi are related by
Vi = sqrt(2 a X)
The cable tension minus the weight is the net force
F = T - M*g = 8683 - 6566 N = 2117 N
Use F = ma to get a and then use the first equation to get Vi.
i got the correct answer but i still get any of the concept >.< i feel dumb ahhh oh oh any good sites to learn physics really good
If you got the correct answer, you must be learning something.
Did you omit the word "don't" after still?
What is the time
about 7
To calculate the initial speed (vi) of the elevator at the beginning of the 6 m descent, we can use the equations of motion.
The equation that relates the final velocity (vf), initial velocity (vi), acceleration (a), and distance (d) is:
vf^2 = vi^2 + 2ad
In this case, the elevator comes to rest at the end of the descent, so its final velocity (vf) is 0 m/s. The acceleration (a) is the acceleration due to gravity (-9.8 m/s^2) because the elevator is decelerating.
We are given the mass of the elevator (m = 670 kg), the distance (d = 6 m), and the tension in the cable (T = 8683 N). We can find the acceleration using Newton's second law of motion:
T - mg = ma
Here, T is the tension (8683 N), m is the mass (670 kg), g is the acceleration due to gravity (9.8 m/s^2), and a is the acceleration. Rearranging the equation to solve for a:
a = (T - mg) / m
Substituting the given values, we have:
a = (8683 N - 670 kg * 9.8 m/s^2) / 670 kg
Now, we can substitute the values of a, d, and vf into the equation relating vf, vi, a, and d:
0^2 = vi^2 + 2 * a * d
Simplifying the equation:
0 = vi^2 + 2 * [(T - mg) / m] * d
0 = vi^2 + 2 * (8683 N - 670 kg * 9.8 m/s^2) / 670 kg * 6 m
Now, solve the equation for vi:
vi^2 = -2 * (8683 N - 670 kg * 9.8 m/s^2) / 670 kg * 6 m
vi = sqrt[-2 * (8683 N - 670 kg * 9.8 m/s^2) / 670 kg * 6 m]
Calculating with the given values will give you the speed vi of the elevator at the beginning of the 6 m descent.