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Citrus2011 [14]
3 years ago
12

Aril does 1800 J of work to get a canoe moving from rest. April and the canoe have a mass of 70 kg.

Physics
2 answers:
wolverine [178]3 years ago
4 0

a) KE=energy required to make them move=1800J

b) KE=1/2*mv^2, so v=\sqrt{\frac{2KE}{m}}=7.17 J

Svetllana [295]3 years ago
4 0

Answer:

a) KE = 1800 J

b) v = 7.17 m/s

Explanation:

Part a)

As we know by work energy theorem that work done by all force is always equal to change in kinetic energy of system

here we have

W = KE_f - KE_i

given that

initial kinetic energy = 0

work done = 1800 J

so final kinetic energy will be

KE = 1800 J

part b)

now we know that

KE = \frac{1}{2}mv^2

now from above equation we have

1800 = \frac{1}{2}(70)v^2

v^2 = \frac{1800}{35}

now from above we have

v = 7.17 m/s

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A 0.29 kg particle moves in an xy plane according to x(t) = - 19 + 1 t - 3 t3 and y(t) = 20 + 7 t - 9 t2, with x and y in meters
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Answer:

Part a)

F = 7.76 N

Part b)

\theta = -137.7 degree

Part c)

\theta = -127.7 degree

Explanation:

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So here we know that

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Part a)

differentiate x and y two times with respect to time to find the acceleration

a_x = \frac{d^2}{dt^2}(-19 + t - 3t^3)

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a_y = -18

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\vec a = (-18t)\hat i + (-18)\hat j

at t= 1.1 s we have

\vec a = -19.8 \hat i - 18 \hat j

now the net force of the object is given as

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\vec F = (0.29 kg)(-19.8 \hat i - 18 \hat j)

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now magnitude of the force will be

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Part b)

Direction of the force is given as

tan\theta = \frac{F_y}{F_x}

tan\theta = \frac{-5.22}{-5.74}

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Part c)

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now at t = 1.1 s

\vec v = -9.89\hat i - 12.8 \hat j

now the direction of the velocity is given as

\theta = tan^{-1}(\frac{v_y}{v_x})

\theta = tan^{-1}(\frac{-12.8}{-9.89})

\theta = -127.7 degree

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