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The velocity of a 1.3 kg block sliding down a frictionless inclined plane is found to be 1.26 m/s. 1.10 s later, it has a velocity of 5.88 m/s.

What is the angle of the plane with respect to tThe velocity of a 1.3 kg block sliding down a frictionless inclined plane is found to be 1.26 m/s. 1.10 s later, it has a velocity of 5.88 m/s.

What is the angle of the plane with respect to the horizontal (in degrees)? he horizontal (in degrees)?

Respuesta :

Answer:

[tex]\theta = 25.3^\circ[/tex]

Explanation:

The acceleration of the block can be found by the kinematics equations:

[tex]v = v_0 + at\\5.88 = 1.26 + a(1.1)\\a = 4.2~m/s^2[/tex]

Since the plane is frictionless, the only force acting on the block along the motion of the block is its weight.

[tex]F = mg\sin(\theta) = ma\\g\sin(\theta) = a\\(9.8)\sin(\theta) = 4.2\\\theta = 25.3^\circ[/tex]

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