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3 years ago

I NEED HELP I AM SO CONFUSED, WILL GIVE BRAIN.

Physics

1 answer:

Phantasy [73]3 years ago

5 0

Answer:

$resolving \: \\ (\downarrow_{ + }) \: s = ut + \frac{1}{2} g {t}^{2} \\ u = 0 ,\: g = 9.81m. {s}^{ - 2} , \: t = 5s \\ s =h= \frac{1}{2} g {t}^{2}, = \frac{1}{2} (9.81){5}^{2}\\=4.9\times25 \\ h=122.6m$

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3 years ago

Determine a formula for the magnitude of the force F exerted on the large block (Mc) so that the mass Ma does not move relative

SVEN [57.7K]

Answer:

The magnitude of the force F is given by

F = ( $M_{a}$ + $M_{b}$ + $M_{c}$ ) *( $M_{b}$ *g/( $\sqrt{M_{a} ^{2}-M_{b} ^{2}}$ ))

Explanation:

Given there are three blocks of masses $M_{a}$ , $M_{b}$ and $M_{c}$ (ref image in attachment)

When all three masses move together at an acceleration a, the force F is given by

F = ( $M_{a}$ + $M_{b}$ + $M_{c}$ ) *a ................(equation 1)

Also it is given that $M_{a}$ does not move with respect to $M_{c}$ , which gives tension T is exerted on pulley by $M_{a}$ only, Hence tension T is

T = $M_{a}$ *a ..........(equation 2)

There is also also tension exerted by $M_{b}$ . There are two components here: horizontal due to acceleration a and vertical component due to gravity g. Thus tension is given by

T = $M_{b}$ $\sqrt{a^{2} +g^{2} }$ ................(equation 3)

From equation 2 and 3, we get

$M_{a}$ *a = $M_{b}$ $\sqrt{a^{2} +g^{2} }$

Squaring both sides we get

$M_{a} ^{2}$ * $a^{2}$ = $M_{b} ^{2}$ * ( $a^{2}$ + $g^{2}$ )

$M_{a} ^{2}$ * $a^{2}$ = ( $M_{b} ^{2}$ * $a^{2}$ )+ ( $M_{b} ^{2}$ * $g^{2}$ )

( $M_{a} ^{2}$ - $M_{b} ^{2}$ ) * $a^{2}$ = $M_{b} ^{2}$ * $g^{2}$

$a^{2}$ = $M_{b} ^{2}$ * $g^{2}$ /( $M_{a} ^{2}$ - $M_{b} ^{2}$ )

Taking square root on both sides, we get acceleration a

a = $M_{b}$ *g/( $\sqrt{M_{a} ^{2}-M_{b} ^{2}}$ )

Hence substituting the value of a in equation 1, we get

F = ( $M_{a}$ + $M_{b}$ + $M_{c}$ ) *( $M_{b}$ *g/( $\sqrt{M_{a} ^{2}-M_{b} ^{2}}$ ))

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3 years ago

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