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Dmitry_Shevchenko [17]
3 years ago
13

A wagon with an initial velocity of 2 m/s and a mass of 60 kg, gets a push with 150 joules of

Physics
1 answer:
Aleks04 [339]3 years ago
7 0

Answer:

v_f = 3 m/s

Explanation:

From work energy theorem;

W = K_f - K_i

Where;

K_f is final kinetic energy

K_i is initial kinetic energy

W is work done

K_f = ½mv_f²

K_i = ½mv_i²

Where v_f and v_i are final and initial velocities respectively

Thus;

W = ½mv_f² - ½mv_i²

We are given;

W = 150 J

m = 60 kg

v_i = 2 m/s

Thus;

150 = ½×60(v_f² - 2²)

150 = 30(v_f² - 4)

(v_f² - 4) = 150/30

(v_f² - 4) = 5

v_f² = 5 + 4

v_f² = 9

v_f = √9

v_f = 3 m/s

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Answer:

F = 318.1 N

Explanation:

As we know that torque to open the nut is given by formula

\tau = \vec r \times \vec F

so we can write it as

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now we know that

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angle between force and the wrench is 41 degree

so we have

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F = \frac{48}{(0.23)sin41}

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