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

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

its a 5.5 n

glad that