Mechanics

The correct solution is:
work done = change of kinetic energy
F x 6 = (1/2)(1000)(10^2)
i.e. F = 8333 N
In your equation, F-mg = ma, is used wrongly.
the acceleration a=8.33 m/s2 is the actual resultant acceleration already taken in consideration the acceleration due to gravity g
therefore, the equation should be written as:
F = m.a = 1000x8.33 N = 8330 N



2007-04-06 14:40:39 補充：
Sorry....I have missed out the weight of the lift in my calculationThe force should be (8333 + mg) = (8333+10000) N = 18333 Nit is because even the lift is stationary, a force of mg(=10000N) is always needed to prevent it from falling.

2007-04-06 14:42:54 補充：
In your equation, you should use F+mg = maif you take upward as positive, notice that a is +ve(acceleration is opposite to velocity, hence become a decceleration).then F+1000x(-10)=1000(8.33), this would give the answer.

2007-04-06 14:43:26 補充：
In your equation, you should use F+mg = maif you take upward as positive, notice that a is +ve(acceleration is opposite to velocity, hence become a decceleration).then F+1000x(-10)=1000(8.33), this would give the answer.

2007-04-07 13:14:18 補充：
taking upward as +vehence, u= -10 m/s , v = 0 m/s, s = -6 m, a=?using v^2-u^2=2.a.s-(-10)^2 = 2.a.(-6)this gives a = +8.33 m/s2i.e. the acceleration is pointing upward, opposite to the direction of lift velocity....this implies a decceleration.

2007-04-07 13:19:23 補充：
Then...use the equation [net force] = m.aF+mg = maF+1000x(-10) = 1000x(+8.33)[remember that upward direction is +ve]thus, F = 1000x(8.33+10) N = 18330 Nthe force is +ve, hence pointing upward.