Aplayground slide is 8.80 ft long and makes an angle of 25.0° with the horizontal. a 63.0-kg child, initially at the top, slides all the way down to the bottom of the slide. (a) choosing the bottom of the slide as the reference configuration, what is the system's potential energy when the child is at the top and at the bottom of the slide? what is the change in potential energy as the child slides from the top to the bottom of the slide? (include the sign of the value in your answer.)

initial: 1654.6 J, final: 0 J, change: -1654.6 J

Explanation:

The length of the slide is

L = 8.80 ft = 2.68 m

So the height of the child when he is at the top of the slide is (with respect to the ground)

The potential energy of the child at the top is given by:

where

m = 63.0 kg is the mass of the child

g = 9.8 m/s^2 is the acceleration due to gravity

h = 1.13 m

Substituting,

At the bottom instead, the height is zero:

h = 0

So the potential energy is also zero: U = 0 J.

This means that the change in potential energy as the child slides down is

the box does not move because you need to use a force greater than 300 n. you are only using 50 n. it needs to at least be 301 n for the box to move because then it is an unbalanced force and it will move the way that the greater force is pushing it. if you use only 300 n it will be a balanced force and will not move because the force you are exerting on the   box is the same as its force.

hope this !

im sorry if i went too in depth, i am going to include a link to a video on khan academy that will explain this.   hope that you like the video and that it explains more in detail to you.

that is the link.

the answer is f1 and d2 because if you put equal force on those two points the beam would balance trust me on this