A 1000 kg elevator accelerates upward at 1.0 m/s2 for 10 m, starting from rest. a. How much work does gravity do on the elevator
? b. How much work does the tension in the elevator cable do on the elevator? c. What is the elevator’s kineti
2 answers:
Answer:
A)Work done by gravity = -98 Kj
B) W_tension = 108 Kj
C) Final kinetic energy Kf = 10 Kj
Explanation:
We are given;
mass; m = 1000kg
Upward acceleration; a = 1 m/s²
Distance; d = 10m
I've attached a free body diagram to show what is happening with the elevator.
A) From the image i attached, the elevators weight acting down is mg.
While T is the tension of the cable pulling the elevator upwards.
Now, we know that,
Work done = Force x distance
Thus, W = F•d
We want to calculate the work done by gravity;
From the diagram I've drawn, gravity is acting downwards and in am opposite direction to the motion having an upward acceleration.
Thus, Force of gravity = - mg
So, Work done by gravity;
W_grav = - mgd
W_grav = -1000 x 9.8 x 10 = -98000 J = -98 Kj
B) Again, W = F.d
W_tension = T•d
Let's find T by summation of forces in the vertical y direction
Thus, Σfy = ma
So, T - mg = ma
Thus, T = ma + mg
T = m(a + g)
Plugging in values,
T = 1000(1 + 9.8)
T = 1000 x 10.8 = 10800 N
So, W_tension = T•d = 10800 N x 10m = 108000 J = 108 Kj
C) From work energy theorem,
Net work = change in kinetic energy
Thus, W_net = Kf - Ki
Where Kf is final kinetic energy and Ki is initial kinetic energy.
Now since the elevator started from rest, Ki = 0 because velocity at that point is zero.
Thus, W_net = Kf - 0
W_net = Kf
Now, W_net is the sum of work done due to gravity and work done due to another force.
Thus, in this case,
W_net = W_grav + W_tension
W_net = -98000 J + 108000 J
W_net = 10000J = 10Kj
So,since W_net = Kf
Thus, Final kinetic energy Kf = 1000J
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