ufffffffffff! get a load of this!?

Question

Q) a winch is used to raise a 200kg load. the maximum power of the winch is 5kW. calculate the greatest possible acceleration of the load when its speed is 2m/s, and the greatest speed at which the load can be raised.

answers: 0.2m/s^2, 2.55m/s

Answer 1

Since Power = dwork/dt
and lifting a mass at constant velocity the work is
dwork=m*g*v*dt
so
dwork/dt=m*g*v
or
v=P/(m*g)
v=2.55 m/s


where a is a function of t
I do't get your answer. I used P=f*v and looked at the instantaneous case.

The power of the motor must do the work of the tension in the cable displaced over a distance dy (note that under constant, or max v, T=m*g)

T-mg=m*a
T=m(g+a)
The work is T*dy
P=m*(g+a)*dy/dt

the instantaneous dy/dt is
v=2 m/s
P/(2*m)-g=a
=5000/(400)-9.81
2.69 m/s^2


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