Physics, 24.03.2020 23:57 esme2583

A 60 g ball is dropped from rest from a height of 2.4 m. It bounces off the floor and rebounds to a maximum height of 1.9 m. If the ball is in contact with the floor for 0.034 s, what is the magnitude (in N) and direction of the average force exerted on the ball by the floor during the collision

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Answer from: mullanebrianot3dpw

The force is 1.34 newtons and its direction is upward.

Explanation:

Choosing positive direction pointing towards the floor in this collision we're going to use the momentum-impulse theorem that states:

$J=\Delta p$ (1)

with $\Delta p=p_f-p_i$ the change in the momentum and J the impulse, with pi the initial momentum that is the momentum just before the collision and pf the final momentum th is the momentum just after the collision. The impulse J is also defined as:

$J=F_{avg}\Delta t$ (2)

with $F_{avg}$ the average force and $\Delta t$ the time the collision lasts

We can equate expressions (2) and (1):

$\Delta p=p_f-p_i=F_{avg}\Delta t$

Using the definition of linear momentum as mass (m) time velocity (v):

$mv_f-mv_i=F_{avg}\Delta t$

We can solve for Favg:

$F_{avg}=\frac{m(v_f-v_i)}{\Delta t}$ (3)

Now we should find the velocities vf and vi, we should do this using conservation of energy:

For the velocity the ball has just before reaches the floor:

U_i=K_f

With Ui the initial potential energy (there is not initial kinetic energy) and Kf the final kinetic energy (there is not final potential energy), then:

$mgh=\frac{mv_i^2}{2}$

solving for vi:

$v_i=\sqrt{2gh}=\sqrt{2*9.81*2.4}=6.86\frac{m}{s}$

For the velocity the ball has just after bounces the floor:

K_i=U_f

There is not initial potential energy because it's a floor level at this instant, and the there is not final kinetic energy because the ball has instantly zero velocity at its maximum height (hm), then:

$\frac{mf_i^2}{2}=mgh_m$

solvig for vf:

$v_f=\sqrt{2gh_m}=\sqrt{2*9.81*1.9}=6.10\frac{m}{s}$

Using vf and vi on (3):

$F_{avg}=\frac{(0.06)(6.10-6.86)}{0.034}=-1.34 N$

The negative sign indicates the direction of the force is pointing away the floor

Answer from: Quest

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.

Answer from: Quest

heats surface water up

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A 60 g ball is dropped from rest from a height of 2.4 m. It bounces off the floor and rebounds to a...

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