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erastovalidia [21]

3 years ago

9

The magnitude and direction exerted by two tugboats towing a ship are 1670 kilograms, N30degreesW, and 1250 kilograms, S60deg

reesW, respectively. Find the magnitude, in kilograms, and the direction angle, in degrees, of the resultant force.

1 answer:

lilavasa [31]3 years ago

8 0

Answer:

magnitude=2085.99 kg

direction angle= 23.18°

Explanation:

let the forces be x,y

x = 1670cos(90°+30°) i +1670sin(90°+30°) j,

y = 1250cos(-90°-60°) i +1250sin(-90°-60°) j

x = 1670cos(120°) i + 1670sin(120°) j,

y = 1250cos(-150°) i + 1250sin(-150°) j

resultant force = x+y

resultant force = 1670cos(120°) i +1670sin(120°) j + 1250cos(-150°) i +1250sin(-150°) j

taking i and j term together

resultant force = (1670cos(120°) +1250cos(-150°)) i +(1670sin(120°) +1250sin(-150°)) j

resultant force = -1917.53 i +821.26j

$magnitude=\sqrt {(-1917.532)^{2} + (821.26)^{2}}$

magnitude=2085.99 kg

direction angle= $tan^{-1}(\frac{821.26}{1917.53})$

direction angle= 23.18°

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Given that,

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Force = 110 N

Angle = 20.2°

Distance = 5.10 m

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(A). We need to calculate the work done by the gravitational force

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Where, w = work

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Put the value into the formula

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(B). We need to calculate the increase in internal energy of the crate-incline system owing to friction

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Put the value into the formula

$\Delta U=-(-158.8)\ J$

$\Delta U=158.8\ J$

(C). We need to calculate the work done by 100 N force on the crate

Using formula of work done

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Put the value into the formula

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We need to calculate the work done by frictional force

Using formula of work done

$W=-f\times d$

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Put the value into the formula

$W=-0.4\times9.2\times9.8\cos20.2\times5.10$

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We need to calculate the change in kinetic energy of the crate

Using formula for change in kinetic energy

$\Delta k=W_{g}+W_{f}+W_{F}$

Put the value into the formula

$\Delta k=-158.8-172.5+510$

$\Delta k=178.7\ J$

(E). We need to calculate the speed of the crate after being pulled 5.00m

Using formula of change in kinetic energy

$\Delta k=\dfrac{1}{2}m(v_{2}^2-v_{1}^{2})$

$v_{2}^2=\dfrac{2\times\Delta k}{m}+v_{1}^2$

Put the value into the formula

$v_{2}^2=\dfrac{2\times178.7}{9.2}+1.58$

$v_{2}=\sqrt{\dfrac{2\times178.7}{9.2}+1.58}$

$v_{2}=6.35\ m/s$

Hence, (A). The work done by the gravitational force is -158.8 J.

(B). The increase in internal energy of the crate-incline system owing to friction is 158.8 J.

(C). The work done by 100 N force on the crate is 510 J.

(D). The change in kinetic energy of the crate is 178.7 J.

(E). The speed of the crate after being pulled 5.00m is 6.35 m/s

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