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2 years ago

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Assume that the cart is free to roll without friction and that the coefficient of static friction between the block and the cart

is μs. Derive an expression for the minimum horizontal force that must be applied to the block in order to keep it from falling to the ground. Express your answer in terms of acceleration of gravity g and some or all of the variables m, M, and μs.

2 answers:

Dahasolnce [82]2 years ago

6 0

Answer: $F=\frac{(M+m)g}{\mu _s}$

Explanation:

Given

Trolley of mass M is free to roll without friction

coefficient of friction between trolley and mass m is $\mu _s$

Force F is applied on mass m

Acceleration of the system is

$a=\frac{F}{M+m}$

friction Force will balance weight of block

friction force $=\mu _sN$

$N=ma$

$\mu _sN=mg$

$\mu _sma=mg$

$\mu _s=\frac{g}{a}$

$F=\frac{(M+m)g}{\mu _s}$

Ainat [17]2 years ago

3 0

Answer:<em> </em>N = <em>fa/g</em>

Explanation:

if the block is not to fall,the friction force, <em>f , </em>must balance the block's weight :

<em>f = mg</em><em>. </em>But the horizontal motion of the block is given by <em>N = ma</em>

Therefore,

<em>f / N = g/ a </em><em> </em>

<em>or </em><em>a = g/f / N</em>

Since the maximum value of <em>f / N is μs</em><em>, </em>we must have <em>a</em> ≥ <em>g/μs,</em><em> </em>if<em> </em>the block is not to fall. Q.E.D

N = <em>fa/g</em>

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A battleship launches a shell horizontally at 100 m/s from the ship’s deck that’s 50 m above the water. The shell is intended to

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The shell will land 10.18m away from the buoy.

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In order to solve this problem, we must first do a sketch of what the problem looks like (see attached picture).

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In order to do so we can use the following equation:

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now, we know that the final height and the initial velocity are to be zero, so we can simplify the equation like this:

$0=y_{0}+\frac{1}{2}at^{2}$

and solve for t:

$t=\sqrt{\frac{-2y_0}{a}}$

now we can substitute the values:

$t=\sqrt{\frac{-2(50m)}{-9.81m/t^2}}$

t=3.19s

Since it takes 3.19s for the shell to hit the water, that's the amount of time it spends flying horizontally.

So we can consider the shell to move at a constant speed if there was no tailwind, so we can find the distance from the ship to point A to be:

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$x_{A}=319m$

We can now find the distance between the ship to point B, which is the point the ball falls due to the tailwind. Since the movement will be accelerated in this scenario, we can find the distance by using the following formula:

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So we can substitute the given values:

$x_{f}=(100m/s)(3.19s)+\frac{1}{2}(2m/s^{2})(3.19s)^{2}$

Which yields:

$x_{f}=329.18m$

so now we can use the A and B points to find by how far the shell missed the buoy:

Distance=329.18m-319m=10.18m

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We have that for the Question "the acceleration of the object at time t = 0.7 s is most nearly equal to which of the following?"

it can be said that the acceleration of the object at time t = 0.7 s is most nearly equal to the slope of the line connecting the origin and the point where the graph where the graph crosses the 0.7s grid line

From the question we are told

the acceleration of the object at time t = 0.7 s is most nearly equal to which of the following?

Generally the equation for the Force is mathematically given as

F=\frac{F}{dx}

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K.E=3J

Therefore

the acceleration of the object at time t = 0.7 s is most nearly equal to the slope of the line connecting the origin and the point where the graph where the graph crosses the 0.7s grid line

For more information on this visit

brainly.com/question/23379286

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