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3 years ago

A woman pulls a 7.87 kg suitcase,

Physics

1 answer:

Lelechka [254]3 years ago

6 0

Answer:

Effective Pulling Force = 29.2(cos 62.7°) = 13.393 N

Resistance to Pulling Force = 11.7 N

Net Force acting on suitcase = 13.393 - 11.7 = 1.693 N

Distance moved = 8.44 m

Net Work = 1.693(8.44) = 14.289 ≈ 14.3 J ANS

Explanation:

hope that makes sense homie

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$u_y = 20.0 m/s$

So we can find the final vertical velocity using the same suvat equation as before:

$v_y^2 - u_y^2 = 2as$

$v_y = \sqrt{u^2+2as}=\sqrt{(20.0)^2+2(-9.8)(-25)}=-29.8 m/s$ (downward, so we take the negative solution)

The horizontal speed instead is zero, since the pebble is initially launched vertically, so the final speed is just equal to the magnitude of the vertical velocity:

v = 29.8 m/s

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This case is similarly to the previous one: the only difference here is that the pebble is launched straight down instead than up, therefore

$u_y = -20.0 m/s$

Using again the same suvat equation:

$v_y^2 - u_y^2 = 2as$

$v_y = \sqrt{u^2+2as}=\sqrt{(-20.0)^2+2(-9.8)(-25)}=-29.8 m/s$ (downward, so we take the negative solution)

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