Work Done by Elevator Lifting Mechanism

[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 I am not getting right answer
whcih is 0.5115MJ
any quick solutoins!
thanks

 

[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?

 

[SS2006]

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

 

[what]

The answer is .5115 MJ
the equation in symbols should look something like this
[tex]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[/tex]
where [tex] F_a_c_c[/tex] is the force exerted by the machinery to accelerate the elevator.