The acceleration that the minimum mass will have is given as: 3.3 [tex](\frac{m}{s^{2}})[/tex]
Acceleration in physics is given as the rate of change of the velocity of an object in relation to time.
On a 20° slope, the force parallel triggers an acceleration down the slope or incline.
This is represented as;
= ∙ ∙ sin = ∙ 9.8 ∙ sin 20°.
The friction force is triggered by the force perpendicular ∙ ∙ cos is
= µ ∙ ∙ cos = 0.81 ∙ ∙ 9.8 ∙ cos 20° .
Because we are attempting to shift the block up the incline, both of these forces are opposing motions: hence
+ = ∙ 9.8 * sin 20° + 0.81 * * 9.8 * cos 20°
= ∙ 9.8(sin 20° + 0.81 * cos 20°)
The force triggered by the 2 kg is
= 2 ∙ 9.8
= 19.6 ( pulling the block up the slope)
The block will remain stationery and will not slip, if the sum of the down forces is below:
19.6 N. * 9.8(sin 20° + 0.81 * cos 20°) < 19.6
m < ([tex]\frac{19.6}{9.8 (sin 20 + 0.81 * cos 20) }[/tex]
The kinetic friction force below is required:
= (µ/ µ) *
= (µ/ µ) µ * * cos
= µ * * cos
= 0.45 * 1.81 * 9.8 * cos 20°.
Hence,
19.6 – 1.81 * 9.8 (sin 20° + 0.45 * cos 20°)
= 1.81 ∙ a
Thus,
a = 3.3 [tex](\frac{m}{s^{2}})[/tex]
Full Question:
Figure shows a block of mass m resting on a 20∘ slope. The block has coefficients of friction 0.81 and 0.45 with the surface. It is connected via a massless string over a massless, frictionless pulley to a hanging block of mass 2.0 kg.
If this minimum mass is nudged ever so slightly, it will start being pulled up the incline. What acceleration will it have?
Learn more about acceleration at:
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