A block is initially at rest on top of an inclined ramp that makes an angle θ0 with the horizontal. The base of the ramp has a length of D . After the block is released from rest, it slides down the ramp onto a rough horizontal surface until it comes to rest at a position x=4D from the base of the ramp, as shown in the figure. There is negligible friction between the block and the inclined ramp, while the coefficient of kinetic friction between the block and the rough horizontal surface is μb .

(a) On the axes below, sketch and label graphs of the following quantities as a function of the position of the block between x=−D and x=4D . Calculations for values for the vertical axis are not necessary, but the same vertical scale should be used for both quantities.
i. The kinetic energy K of the block
ii. The gravitational potential energy Ug of the block-Earth system

(b) The block is released from the top of a new ramp that has a base length of 2D, but still makes an angle θ0 with the horizontal. A student is asked to predict whether the final horizontal position of the block will be twice as far from the base of the ramp compared to when it was released from the original ramp. The student reasons that since the block will be released from a new height that is twice as high as the original height, the block will have more energy when it reaches the base of the ramp, so it will slide farther along the right surface before stopping at a position x=8D.
i. Which aspects of the student’s reasoning, if any, are correct? If no aspect of the student’s reasoning is correct, write “none”.
ii. Which aspects of the student’s reasoning, if any, are incorrect? If no aspect of the student’s reasoning is incorrect, write “none”.

(c) Derive an equation for the new final position of the block. Express your answer in terms of D.

(d) In the following question, refer to the relationships written in part (c), not just the final answer obtained by manipulating those relationships.
For any correct aspects of the student’s reasoning identified in part (b)(i), how is the student’s reasoning expressed by your mathematical relationships in part (c) ?
For any incorrect aspects of the student’s reasoning identified in part (b)(ii), how do your relationships in part (c) correct the student’s incorrect reasoning?

Question

khalil.waters · Accepted Answer

a) i) See graph in attachmentii) See graph in attachment.b) i) All of them ii) nonec) [tex]s=\frac{4D tan 	heta}{\mu_s}[/tex]d) See explanation belowExplanation:a)Find in attachment the graph showing the kinetic energy and the potential energy versus the position, x.i)Between x = -D and x = 0, the block is sliding down along the ramp. The kinetic energy of the block at any time is given by:[tex]KE=\frac{1}{2}mv^2[/tex]wherem is the mass of the blockv is its speedAt the beginning, the block's kinetic energy is zero, because the speed is initially zero: since v = 0, KE = 0.As the block slides down, the kinetic energy increases, because the speed of the block increases; at x = 0 (end of the ramp), all the initial energy has been converted into kinetic energy, which is now maximum.Then, the block slides along the flat, rough surface; as friction does (negative) work on the block, the speed of the block decreases, and so also its kinetic energy decreases, becoming zero when x = +4D (when the block comes to a stop).ii)The potential energy of the block is given by:[tex]GPE=mgh[/tex]wherem is the mass of the blockg is the acceleration due to gravityh is the height of the block above the groundAt the beginning (x = -D) the potential energy is maximum since the block is at maximum height.When the block slides down (between -D and 0), the height h decreases, therefore the potential energy decreases as well, until reaching 0 when x = 0 (end of the ramp).After x = 0, the block slides along the rough surface; however, its potential energy here no longer changes, as the height h dors not change (the surface is horizontal).b)Here, the block is released from the top of a new ramp, which has a base length of 2D (instead of D) but same angle as before: therefore, the initial height of the ramp is twice that in part a). This also means that the initial (potential) energy of the block in this case is twice the GPE of part a):[tex]GPE'=2GPE[/tex]As a result, when the block reaches the end of the ramp at x = 0, it will have twice the kinetic energy it had before:[tex]KE'=2KE[/tex]The stopping distance of an object moving with accelerated motion is proportional to its initial kinetic energy:[tex]s\propto KE[/tex]Therefore, this means that here the stopping distance of the block will be twice that of part a (which was 4D), so the block will stop at x = +8D.So, all aspects of the student's reasoning are correct.c)Let's call [tex]E[/tex] the initial total energy of the block at the top of the ramp.In situation a), the initial total energy is[tex]E=GPE=mgh = mgD tan 	heta[/tex]where [tex]h=Dtan 	heta[/tex] is the height of the ramp.And so the kinetic energy at the bottom of the ramp is[tex]KE=E[/tex]We can rewrite the kinetic energy so that[tex]\frac{1}{2}mv^2=E ightarrow v \sqrt{\frac{2E}{m}}\ightarrow v=\sqrt{2gD tan 	heta}[/tex] For an accelerated motion, the stopping distance can be found using the equation[tex]v'^2-v^2=2as[/tex]where[tex]v'=0[/tex] is the final speed of the block[tex]a=-\mu_b g[/tex] is the acceleration due to frictionSo we find[tex]s=\frac{-v^2}{2a}=\frac{(2gD tan 	heta)}{\mu_s g}=\frac{2D tan 	heta}{\mu_s}[/tex] (1)In situation b), the initial height of the block is[tex]h=2D tan 	heta[/tex]So the final stopping distance becomes (1)[tex]s=\frac{4D tan 	heta}{\mu_s}[/tex]d)We can see that the formula derived in part c) depends only on:- The initial height of the ramp, which is [tex]Dtan 	heta[/tex] in part a) and [tex]2D tan 	heta[/tex] in part b)- The coefficient of friction in the rough part, [tex]\mu_s[/tex]- The angle of the ramp, which remains the same in the two casesTherefore, all the correct aspects identified by the student in his reasoning are found in the fact that the final stopping distance is proportional to the initial energy of the block, which is proportional to initial height of the block, and since this is twice in part b) compared to part a), therefore the stopping distance is also twice in part b).

bkovacek · Answer

Refer the below solution for better understanding.

Given :

Block has initial velocity, u = 0.

Base of the ramp has a length of D.

Negligible friction between the block and the inclined ramp.

Coefficient of kinetic friction between the block and the rough horizontal surface is [tex]\rm \mu_ b[/tex].

Solution :

a) Graph of the following quantities is attached below.

i) We know that the Kinetic energy is,

[tex]\rm KE = \dfrac{1}{2}mv^2[/tex]

at x = 0, KE is zero because initial velocity is zero but kinetic energy increases as velocity increases. And at an instant kinetic energy becomes maximum because velocity is maximum.

When the block slides along the flat, rough surface, friction acts on the block, the speed of the block decreases, so kinetic energy is also decreases and becomes zero at x = 4D.

ii) We know that potential energy is given by,

PE = mgh

At x = -D , height of the block is maximum therefore potential is also maximum at x = -D. The potential energy decreases as well, until reaching 0 when x = 0.

b) The initial height of the ramp is twice that in part a). This also means that the initial (potential) energy of the block in this case is twice the PE of part a). As a result, when the block reaches the end of the ramp at x = 0, it will have twice the kinetic energy it had before.

Stopping distance is proportional to its initial kinetic energy:

Therefore, the stopping distance of the block will be twice that of part a), so the block will stop at x = 8D.

Therefore, all aspects of the student's reasoning are correct.

c) Initial total energy is,

E = PE

[tex]\rm E= mgh = mgDtan\theta[/tex]

Kinetic energy at bottom of the ramp is,

KE = E

[tex]\rm \dfrac{1}{2}mv^2 = mgDtan\theta[/tex]

[tex]\rm v = \sqrt{2gDtan\theta}[/tex]

We know that,

[tex]\rm v'^2 = u^2 +2as[/tex]

here,

[tex]\rm v' = 0[/tex] and [tex]\rm a = -\mu_bg[/tex] (acceleration due to friction). So,

[tex]\rm s = \dfrac{-u^2}{2a}[/tex]

[tex]\rm s = \dfrac{2gDtan\theta}{\mu_bg}[/tex]

[tex]\rm s = \dfrac{2Dtan\theta}{\mu_b}[/tex] ----- (1)

In section b) h = [tex]\rm2Dtan\theta[/tex]

Now equation (1) becomes,

[tex]\rm s = \dfrac{4Dtan\theta}{\mu_b}[/tex]

d) We can see that the formula derived in part c) depends only on:

The initial height of the ramp, which is [tex]\rm Dtan\theta[/tex] in part a) and [tex]\rm 2Dtan\theta[/tex] in part b).

The coefficient of friction in the rough part, [tex]\mu _s[/tex].

The angle of the ramp, which remains the same in the two cases.

Therefore, all the correct aspects identified by the student in his reasoning are found in the fact that the final stopping distance is proportional to the initial energy of the block, which is proportional to initial height of the block, and since this is twice in part b) compared to part a), therefore the stopping distance is also twice in part b).

For more information, refer the link given below

https://brainly.com/question/999862?referrer=searchResults

A Block Is Initially At Rest On Top Of An Inclined Ramp That Makes An Angle 0 With The Horizontal. The

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