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1) The initial speed of the block is $(3,0)$ with respect to the ground 2) The initial position of the block is $(0,0)$ with respect to the ground 2) The speed of the platform is $(0,4)$ with respect to the ground 3) The friction force has a magnitude equal to the $\mu \, m \, g$, and a direction opposite to the relative velocity of the block with respect to the platform. Both of these velocities are given with reference to ground. In other words:

Running the simulation over time results in the block coming to rest relative to the platform after $1.666$ seconds, at which time the $x$ coordinate of the block is $2.5$ with respect to ground, and the $y$ coordinate is $3.333$ with respect to ground. The block is a distance $4.1666$ from the origin, and it has moved a total arc length of $4.753$ with respect to ground.

The $(x,y)$ trajectory of the block relative to ground is shown below:

If that is the case, my last paragraph changes as follows:

Running the simulation over time results in the block coming to rest relative to the platform after $1$ second, at which time the $x$ coordinate of the block is $1.5$ with respect to ground, and the $y$ coordinate is $4$ with respect to ground. The block is a distance $4.272$ from the origin, and it has moved a total arc length of $4.348$ with respect to ground.

@Kudo Shinichi
–
I wanted to check my original solution, since it was a bit involved. So then I remembered the first postulate of special relativity: "The laws of physics are the same in all inertial frames of reference". Since the platform is not accelerating, an observer sitting on the platform and watching should observe "ordinary" physics. And we can apply simple kinematics to that case. Let $v_0$ be the initial velocity of the block relative to the platform. There is a constant acceleration relative to the platform equal to $\mu \, g$.

$\frac{v_0}{\mu \, g} = t_f \\
t_f = 1$

Then the distance the block slides relative to the platform is:

And over the one second, the platform has moved 4 meters. So the final position of the block with respect to the ground is $(\frac{3}{2}, 4)$. This matches the more complicated approach I took earlier.

Then, to find the total arc length traveled with respect to ground, begin with the expressions for the velocity relative to ground.

Yes........I mean, I solved it using relative velocity.......View the motion of the block with respect to the horizontal platform........The time after which it stops is 1.66666 seconds.....I think you can proceed from here onwards....:)

@Aaghaz Mahajan bro but then from that frame answer I am getting is wrong , can u tell what r u getting ?? Its difficult for me to get answer using relative frame I used that.
.

@Kudo Shinichi
–
Well, what is the correct answer??? I am getting the answer to be 5 m.........Here is my method:- Taking the velocity of block wrt to the platform, we get it to be <3,-4> i.e. v = 5 m/s. Also, the acceleration is -3 m/s^2. So, the time taken is 1.66666666 seconds. Now, the displacement of the block in that time is <5,-20/3> and the displacement of the platform is <0,4>. Using relative velocity again, we get the displacement of the block wrt ground to be <5,0>......

1) I interpret the "4" in your energy expression as being the final velocity with respect to ground, which is the same velocity as the platform. It's important to note that it is relative to ground.

2) The "3" in your energy expression is relative to the platform (per your previous clarification). So you have two different references in your energy equation, which seems problematic.

3) The other concern is that the work calculation has the following general form:

$W = \int \vec{F} \cdot \vec{d \ell}$

If the force magnitude is constant AND the force is everywhere parallel to the motion, this reduces to:

$W = |\vec{F}| \, \Delta \ell$

How do you know that these simplifications should occur?

$\vec v_b=3 \vec{i}, \vec{a}= \dfrac{3(-3 \vec i +4 \vec j)}{5}\, |\vec a|=3, |\vec v_{b/p}|=5 \implies t_0=5/3 s.$ (Using formula $0-5=-3t_0$)
For block, $\vec{s}_b=3t_0 \vec i + (1/2) \vec{a} {t_0}^2$
Use this to get $\vec{s}=5 \vec i + \dfrac{10}{3} \vec j$
Note: $v_b$ is velocity of block in ground frame, $v_{b/p}$ is velocity of block in platform frame.

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## Comments

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TopNewestHere is my solution

Initial assumptions:

1)The initial speed of the block is $(3,0)$ with respect to the ground2)The initial position of the block is $(0,0)$ with respect to the ground2)The speed of the platform is $(0,4)$ with respect to the ground3)The friction force has a magnitude equal to the $\mu \, m \, g$, and a direction opposite to the relative velocity of the block with respect to the platform. Both of these velocities are given with reference to ground. In other words:$\vec{F}_{friction} \propto \, -(\vec{v}_{block} - \vec{v}_{platform})$

Running the simulation over time results in the block coming to rest relative to the platform after $1.666$ seconds, at which time the $x$ coordinate of the block is $2.5$ with respect to ground, and the $y$ coordinate is $3.333$ with respect to ground. The block is a distance $4.1666$ from the origin, and it has moved a total arc length of $4.753$ with respect to ground.

The $(x,y)$ trajectory of the block relative to ground is shown below:

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Sir initial speed of block should be wrt platform . As such in ground frame it has initial speed to be net 5

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If that is the case, my last paragraph changes as follows:

Running the simulation over time results in the block coming to rest relative to the platform after $1$ second, at which time the $x$ coordinate of the block is $1.5$ with respect to ground, and the $y$ coordinate is $4$ with respect to ground. The block is a distance $4.272$ from the origin, and it has moved a total arc length of $4.348$ with respect to ground.

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@Steven Chase Sir I have posted my energy method above with the question.

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@Steven Chase Sir can u pls tell.is there anything wrong I did ?

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$v_0$ be the initial velocity of the block relative to the platform. There is a constant acceleration relative to the platform equal to $\mu \, g$.

I wanted to check my original solution, since it was a bit involved. So then I remembered the first postulate of special relativity: "The laws of physics are the same in all inertial frames of reference". Since the platform is not accelerating, an observer sitting on the platform and watching should observe "ordinary" physics. And we can apply simple kinematics to that case. Let$\frac{v_0}{\mu \, g} = t_f \\ t_f = 1$

Then the distance the block slides relative to the platform is:

$D = \frac{1}{2} \, \mu \, g \, t_f^2 = \frac{3}{2}$

And over the one second, the platform has moved 4 meters. So the final position of the block with respect to the ground is $(\frac{3}{2}, 4)$. This matches the more complicated approach I took earlier.

Then, to find the total arc length traveled with respect to ground, begin with the expressions for the velocity relative to ground.

$(v_x,v_y) = (3 - 3 t, 4) \\ v = \sqrt{25 - 18t + 9 t^2}$

The total arc length is:

$L = \int_0^1 \sqrt{25 - 18t + 9 t^2} \, dt = 4.3484$

This also matches the result of my previous solution

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@Steven Chase sir please help with this problem discussion on a David Morin Question.

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Can it be solved without energy method ? @Steven Chase sir @Aaghaz Mahajan bro ??

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Yes........I mean, I solved it using relative velocity.......View the motion of the block with respect to the horizontal platform........The time after which it stops is 1.66666 seconds.....I think you can proceed from here onwards....:)

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@Aaghaz Mahajan bro but then from that frame answer I am getting is wrong , can u tell what r u getting ?? Its difficult for me to get answer using relative frame I used that. .

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@Aaghaz Mahajan bro. We need total distance not displacement

Answer is 7/6 m onlyLog in to reply

@Steven Chase Sir u r method is correct , but what's incorrect in my method (energy)??

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1)I interpret the "4" in your energy expression as being the final velocity with respect to ground, which is the same velocity as the platform. It's important to note that it is relative to ground.2)The "3" in your energy expression is relative to the platform (per your previous clarification). So you have two different references in your energy equation, which seems problematic.3)The other concern is that the work calculation has the following general form:$W = \int \vec{F} \cdot \vec{d \ell}$

If the force magnitude is constant AND the force is everywhere parallel to the motion, this reduces to:

$W = |\vec{F}| \, \Delta \ell$

How do you know that these simplifications should occur?

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Which dimension is gravity in (i, j, or k)?

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Sir, it is written that x-y plane is a horizontal plane, so I think g should be in k.....

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@Steven Chase sir u got the answer? By other method?

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$\vec v_b=3 \vec{i}, \vec{a}= \dfrac{3(-3 \vec i +4 \vec j)}{5}\, |\vec a|=3, |\vec v_{b/p}|=5 \implies t_0=5/3 s.$ (Using formula $0-5=-3t_0$) For block, $\vec{s}_b=3t_0 \vec i + (1/2) \vec{a} {t_0}^2$ Use this to get $\vec{s}=5 \vec i + \dfrac{10}{3} \vec j$ Note: $v_b$ is velocity of block in ground frame, $v_{b/p}$ is velocity of block in platform frame.

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