# Work Question

1. Oct 27, 2005

### SS2006

A 2000kg elevator rises from rest in the basemtn to the fourth floor, a distance of 25m. As it passes the fourth floor, its speed is 3.0 m/s. There is a constant frictional force of 500N. Calculate the work done by the lifting mechanism.

I tried doing v2square = v1square + 2ad and i get 0.18 acceleration
plugged that into f=ma
to get force,
did force X distance + (friction X distance)
and im not getting right answer
whcih is 0.5115MJ
any quick solutoins!
thanks

2. Oct 27, 2005

### what

Are you sure you are taking into account all the forces acting on the elevator...you neglected the force of gravity, see the elevator needs to do work to surpass the force of gravity, the frictional force and to accelerate the elevator, can you go from there?

3. Oct 27, 2005

### SS2006

i tried what u jsut said, are you getting 0.5115 SPOT ON
im getting like 0.505 and 0.522

4. Oct 28, 2005

### what

$$F_g_r_a_v_i_t_y * d + F_f_r_i_c_t_i_o_n * d + F_a_c_c *d = .5115MJ$$
where $$F_a_c_c$$ is the force exerted by the machinery to accelerate the elevator.