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2 years ago

Calculate the amount of work done if you use a 100N force to push a 50kg box 5m across the kitchen floor.

2 answers:

6 0

So if the formula for work is force times displacement times cosine(theta), you'd plug in the numbers

100x5 (since there's no angle in the problem, cosine(theta) isn't used

100x5 = 500

So the answer would be B.

Hope that helps!

Naya [18.7K]2 years ago

3 0

We know, W = F * s
Here, F = 100 N
s = 5 m

Substitute their values,
W = 100 * 5
W = 500 J

In short, Your Answer would be Option B

Hope this helps!

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By definition of average velocity,

$\vec v_{\rm ave} = \dfrac{\Delta \vec r}{\Delta t} = \dfrac{(x_2-x_1)\,\vec\imath + (y_2-y_1)\,\vec\jmath}{10.7\,\mathrm s}$

$\vec v_{\rm ave} = (-2.25\,\vec\imath + 1.79\,\vec\jmath)\dfrac{\rm m}{\rm s}$

and $x_1=-17.5\,\mathrm m$ and $y_1=23.1\,\mathrm m$ , then

$-2.25\dfrac{\rm m}{\rm s} = \dfrac{x_2 - (-17.5\,\mathrm m)}{10.7\,\mathrm s} \\\\ 1.79\dfrac{\rm m}{\rm s} = \dfrac{y_2 - 23.1\,\mathrm m}{10.7\,\mathrm s}$

Solve for $x_2$ and $y_2$ :

$x_2 = 17.5\,\mathrm m + \left(-2.25\dfrac{\rm m}{\rm s}\right)(10.7\,\mathrm s) \approx \boxed{-6.58\,\mathrm m}$

$y_2 = 23.1\,\mathrm m + \left(1.79\dfrac{\rm m}{\rm s}\right)(10.7\,\mathrm s) \approx \boxed{42.3\,\mathrm m}$

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2 years ago

Question 1 of 10

kirill115 [55]

Answer:

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3 years ago

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Option A

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Hence, option A is correct

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3 years ago