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4 years ago

13

A person rolls a barrel up an inclined plane with a length of 6m and a height of 3m/ How much force is needed to roll up the bar

rel the length of the inclined plane if the barrel weighs 500N? How much work is done on the barrel?

1 answer:

snow_lady [41]4 years ago

8 0

Answer:

Length (l) = 6 m

height (h) = 3 m

Load(L) = 500 N

Effort (E) = ?

we know the principal that

E * l = L * h

6 E = 500 * 3

6E = 1500

E = 250

therefore 250 N work is done on the barrel.

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Read 2 more answers

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Part 2)

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Part 3)

$L = 663.07 kg m^2/s^2$

Part 4)

$\omega = 1.83 rad/s$

Part 5)

$F_c = 453.6 N$

Explanation:

Part a)

Initial angular momentum of the merry go round is given as

$L_1 = I \omega$

here we know that

$I = 194 kg m^2$

$\omega = 1.47 rad/s^2$

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$L_1 = 194 \times 1.47$

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Part b)

Angular momentum of the person is given as

$L = mvR$

so we have

$m = 68 kg$

$v = 4.9 m/s$

R = 1.99 m

so we have

$L_2 = (68)(4.9)(1.99)$

$L_2 = 663.07 kg m^2/s^2$

Part 3)

Angular momentum of the person is always constant with respect to the axis of disc

so it is given as

$L = 663.07 kg m^2/s^2$

Part 4)

By angular momentum conservation of the system we will have

$L_1 + L_2 = (I_1 + I_2)\omega$

$185.2 + 663.07 = (194 + 68(1.99^2))\omega$

$848.27 = 463.28 \omega$

$\omega = 1.83 rad/s$

Part 5)

Force required to hold the person is centripetal force which act towards the center

so we will have

$F_c = m\omega^2 R$

$F_c = 68(1.83^2)(1.99)$

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