Time for 120W motor to lift 20 kg load to 0.5 m/s

[Grindelwald]

Okay, here is a question I just can't solve.

The 120W motor starts to lift a load of 20 kg. During which time, this load will reach a speed of 0.5 m / s, taking into account the potential energy.
PS: Ignore losses in the mechanism!

 

[Baluncore]

Welcome to PF.
Is this question homework?

You know that potential energy; Ep = m·g·h
How much energy will be used to maintain a vertical velocity of 0.5 m/s ?
Is that more or less than the motor 120 W ?

As the mass starts to move slowly, all 120 W will be available to accelerate the mass.
Ek = ½·m·v²

What is the actual question.

 

[Grindelwald]

Welcome to PF.
Is this question homework?

You know that potential energy; Ep = m·g·h
How much energy will be used to maintain a vertical velocity of 0.5 m/s ?
Is that more or less than the motor 120 W ?

As the mass starts to move slowly, all 120 W will be available to accelerate the mass.
Ek = ½·m·v²

What is the actual question.

Thank you for welcoming me.

Yes it is.

Alright, I know formula for potential energy.
Motor power remains the same. 120 W.

Question is how much time is needed to reach the speed of 0.5 m/s by lifting mass of 20 kg with motor 120 W taking in count potential energy. I hope you can understand me because my English knowledge is not that good.

 

[anorlunda]

[Moderator: Moved from a technical forum. No template.]

 

[jack action]

This is an energy conservation problem.

The energy $dE$ provided for a constant power $P$ applied during time interval $dt$ is $dE = Pdt$.

The energy required for a force $mg$ applied during a distance $dx$ is $dE_p = mgdx$.

The energy required for a force $ma$ applied during a distance $dx$ is $dE_k = madx$.

Therefore, what you are providing should equal the sum of what is required. $dt$ is what your looking for. What you know is the velocity $dv$ (I'm assuming the initial velocity is 0), not the distance $dx$. So you have to transform the expression on the 'required' side in terms of $v$, $dv$, $t$ and/or $dt$. Solve the obtained differential equation to find $\Delta t$.