A 562 N truck slides down a frictionless plane inclined at an angle of 30.0 degrees from the horizontal. A Find the acceleration

Question

A 562 N truck slides down a frictionless plane inclined at an angle of 30.0 degrees from the
horizontal.
A Find the acceleration of the trunk.
B Find the acceleration of the trunk if the coefficient

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niczorrrr 3 years 2021-08-08T12:32:29+00:00 1 Answers 152 views 0

Answers ( )

  1. Verity
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    2021-08-08T12:33:36+00:00
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    A) 4.9 m/s^2

    B) 2.4 m/s^2

    Explanation:

    A)

    When the truck is sliding down the frictionless ramp, the only force acting along the direction parallel to the ramp (downward) is the component of the weight of the truck parallel to the ramp, which is given by

    F=mg sin \theta

    where

    mg=562 N is the weight of the truck

    \theta=30.0^{\circ} is the angle of the ramp

    So the net force on the truck is

    F=(562)(sin 30^{\circ})=281 N

    The mass of the truck is given by

    m=\frac{mg}{g}=\frac{562}{9.8}=57.3 kg

    where g=9.8 m/s^2 is the acceleration due to gravity.

    Therefore, the acceleration of the truck along the ramp can now be found using Newton’s Second Law of motion:

    a=\frac{F}{m}=\frac{281}{57.3}=4.9 m/s^2

    B)

    In this case, there is another force acting on the truck: the  force of friction.

    This force acts in the direction opposite to the component of the weight along the plane (so, up along the ramp), and its magnitude is

    F_f =\mu mg cos \theta

    where

    mg=562 N is the weight of the truck

    \theta=30.0^{\circ} is the angle of the ramp

    \mu = 0.30 is the coefficient of friction between the truck and the ramp

    Therefore, the net force acting on the truck in this second case is:

    F=mg sin \theta - \mu mg cos \theta = (562)(sin 30^{\circ})-(0.30)(562)(cos 30^{\circ})=135.0 N

    Therefore, since the mass of the truck is

    m = 57.3 kg

    We can find the new acceleration of the truck:

    a=\frac{F}{m}=\frac{135}{57.3}=2.4 m/s^2

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