Work Done by Elevator Lifting Mechanism

AI Thread Summary
The discussion focuses on calculating the work done by an elevator lifting mechanism as it moves a 2000kg elevator 25m with a final speed of 3.0 m/s, while overcoming a constant frictional force of 500N. Participants emphasize the importance of accounting for all forces, including gravitational force, friction, and the force required for acceleration. The correct approach involves using the equation that combines gravitational force, friction, and acceleration forces multiplied by the distance traveled. Despite attempts to solve the problem, participants report varying results, with the correct answer being 0.5115 MJ. The conversation highlights the necessity of thorough force analysis in physics problems.
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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
 
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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?
 
i tried what u just said, are you getting 0.5115 SPOT ON
im getting like 0.505 and 0.522
 
The answer is .5115 MJ
the equation in symbols should look something like this
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.
 
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