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A 45 kg wagon is being pulled with a rope that makes an angle of 38 degrees with the horizontal. The applied force is 410 N and the coefficient of kinetic friction between the wagon and the ground is 0.18.

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Answer:

7.35m/s²

Step-by-step explanation:

From the question we are not told what to find but we can as well find the acceleration of the wagon.

According to newton second law of motion


\sum F_x = ma_x\\Fm - Ff = ma_x\\Fm - \mu R = ma_x\\Fm- \mu mg = ma_x\\

Fm is the moving force = 410N


\mu is the coefficient of friction = 0.18

m is the mass = 45kg

g is the acceleration dur to gravity = 9.8m/s²

a is the acceleration of the wagon

Substitute the given data into the equation ang get ax


Fm- \mu mg = ma_x\\410 - (0.18)(9.8)(45) = 45a_x\\410 - 79.38 = 45a_x\\330.62 = 45a_x\\a_x = 330.62/45\\a_x = 7.35m/s^2\\

Hence the acceleration of the wagon is 7.35m/s²

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