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Reptile [31]
3 years ago
14

A rubber toy duck is at rest on an inclined plane. When the angle of inclination of the plane is increased to 36.0°, the toy duc

k begins to slide down the incline. Then, when the angle is decreased to 29.0°, the speed of the toy duck is constant. What are the coefficients of static and kinetic friction between the toy duck and the incline?

Physics
1 answer:
ch4aika [34]3 years ago
3 0

Answer

toy begins to slide when the plane is at an angle of 36.0°

at this point the toy has just overcome the static friction

f_s = \mu_s N

f_s = \mu_s mgcos \theta

since toy begins to slide

f_s = mgsin \theta

now on comparing

mgsin \theta = \mu_s mgcos \theta

\mu_s = tan \theta

\mu_s = tan (36^0)

\mu_s = 0.726

as the body is moving at constant velocity acceleration will be zero

so now,

m g sin \theta - f_k = ma

m g sin \theta - \mu_k mg cos\theta = 0

\mu_k = tan \theta

\mu_k = tan (29^0)

\mu_k = 0.554

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Answer:

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Explanation:

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Final displacement, Δd = 2.8·\hat j + 1.0·\hat i +(-1.6·\hat j) = 1.2·\hat j + 1.0·\hat i

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Where;

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The magnitude of the resultant displacement vector is given as follows

\left | d \right | = √((Δx)² + (Δy)²) = √(1² + 1.2²) ≈ 1.6 (To the nearest tenth)

The magnitude of the resultant displacement vector ≈ 1.6 km

The direction of the resultant vector is positive for both the east and north direction, therefore, the direction of the resultant vector = NE

Therefore, the resultant displacement of the delivery truck is approximately 1.6 km, NE from the origin.

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Explanation:

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