Power, force, velocity and tension...

[hello478]

We don't care about whaat causes the motion.

You want to find speed of the object using ##P=\vec F. \vec v## and you have power of the Tension force so that F in above equation represents tension force.

why dont we care about it? how are we supposed to find the velocity then?

 

[BvU]

so which force am i supposed to use to calculate the velocity?

The ##F## in ##P = \vec F \cdot \vec v## is the one that does the work. The motor does the work and the only way it is exerting a force is in the form of tension in the cable. The ##T## you calculated in a .

why dont we care about it? how are we supposed to find the velocity then?

We don't care about what causes the motion: we are given that the speed is constant (##\Rightarrow F_{net}=0##) and we know the opposing force component along the slope (##-mg\sin\theta## -- the minus sign is because its direction is to the left. g = 9.8 m/s²). So we know ##F## in #33 (aka The ##T## you calculated in a ).

(as you can see I also didn't bother to draw more than just ##F## in #33)

##\ ##

 

[MatinSAR]

why dont we care about it? how are we supposed to find the velocity then?

Earlier you've said :

yes, so then i would use the tension force and multiply with velocity, because it is the force causing upward motion? is this explanation correct?

And I disagree. You should use the force that is related to the motor power. It doesn't necessary for the force to cause upward/downward motion.

My interpretation of the question is apparently different from @haruspex . I can't understand why in his opinion the force is in the downslope direction. I think it is better for you to use the ideas of @BvU and @haruspex . I prefer not to contribute anymore. This way you don't get misled by different ideas.

 

[hello478]

ok i sort of get it now... thank you very much!