AS Physics Mechanics doubts

-A person weighing 100 N stands on some bathroom scales in a lift. If the scales show a
reading of 110 N, which answer could describe the motion of the lift?
Can someone explain why the answer is ""Moving downwards and decelerating.""?

For Question 1, in which way does tension act and thus, how is it resolved?
In general, does tension act in the opposite direction to the rope being stretched?

How do I go about sketching the graph in Question 2 ?

And for Question 3, Im aware that i have to use gh=(1/2)(v^2) but what height do I take? The mark scheme say's 0.6m and I'm not sure why i must use this height..

Finally, for Question 4 (part c ) why is the resistive force the horizontal component of force? Shouldn't it be a force that acts in the opposite direction to the moving sled?

Let's take up as positive. Then the net force of the person, which is $ma$ is equal to the the normal reaction force of the floor minus weight. (Normal reaction force is equal in magnitude to the reading of the bathroom scales). (1) Try to figure out whether the acceleration is positive (up) or negative (down). (2) In what situations is it possible?




Yeah, this is more or less correct. Tension is a reaction force. It always acts in the direction of the rope, opposing the force that is exerted on its end tending to stretch it.




Well, there is no question asked in your attachment. Only a graph is there.




Height h here is the difference in height between the final and initial positions. It's because the change in potential energy is $\Delta U=mgh_1-mgh_0=mg\Delta h$. And the law of conservation of energy says that $\mathrm{(change\ in\ potential\ energy)+(change\ in\ kinetic\ energy)=0}$.




You're right about the frictional force opposing movement. However, in the first sentence it says "The child and sledge are pulled across level ground [...]". Doesn't it imply that they're moving on a horizontal plane?

Thanks for the reply!
and I meant for question to 2 : How do I go about sketching the speed time version of the graph?

Oh, I see.

Velocity is the derivative of displacement with regard to time:

$\mathbf{v}=\dfrac{\mathrm{d} \mathbf{r}}{\mathrm{d} t}$.

Therefore if you have a distance-time graph, plot its derivative versus time to obtain the speed-time graph.

What if I was not supposed to use calculus?

Edit: Apologies if Jaroc was getting to this, was posted two days ago so I assumed replies would be quite slow.