An elevator of mass 1000 kg is pulled upwards by a steel cable; the tension in t

Question

An elevator of mass 1000 kg is pulled upwards by a steel cable; the tension in the cable is 11000N. The elevator slides on vertical rails which exert a frictional force of 500 N. The elevator starts from rest and increases its altitude by 20 m. Use g = 10 m s-2 . a. Calculate the work done by the tension force ________________________ J friction ________________________ J, and gravity _________________________ J. What is the total work done on the elevator? b. What is the change in potential energy of the elevator? _______________J c. Use the answers to parts (a) and (b) to calculate the change in kinetic energy of the elevator and hence its speed at the 20 m altitude d. Use this to calculate the acceleration of the elevator

Explanation / Answer

Given mass of elevator m = 1000 kg

tension T = 11000 N

frictional force f = 500 N

displacement s = 20 m

here tension is in upward so taken as positive

friction and gravitational force are in downward so taken as negative

The definition of work is W = force * displacement

a)

W(T) = T * s = 11000 * 20

W(T) = 220000 J

work done by friction

W(f) = - f * s = - 500 * 20 = - 10000 J

work done by gravity is

W(g) = - m * g * s = - 1000 * 9.8 * 20

W(g) = 196000 J

total work W = W(T) + W(f) + W(g)

W = 220000 - 10000 - 196000

W = 14000 J

b)

the change in potential energy is P.E = m * g * h

P.E = 1000 * 9.8 * 20

P.E = 196000 J

c) the total work done = change in kinetic energy

Change in K.E = 14000 J

1/2 * m * v^2 = 14000

1/2 * 1000 * v^2 = 14000

v = 5.29 m/s

d)

from the equation

v^2 - u^2 = 2 * a * s

(5.29)^2 - 0 = 2 * a * 20

a = 0.7 m/s^2

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