Asked by Genny
A 54.0kilogram block is placed on a plane that is inclined 32 degrees to the horizontal. The coefficient of static friction between the block and the plane is 0.573 and the coefficiencet of kinetic friction is 0.310. Find the acceleration of the block down the plane.
Please walk me through your steps instead of just suggesting a formula..Thank you
Please walk me through your steps instead of just suggesting a formula..Thank you
Answers
Answered by
Henry
Wb = m*g = 54kg * 9.8N/kg = 529.2 N. =
Wt of the block.
Fb = 529N @ 32 Deg. = Force of block.
Fp = 529*sin32 = 280.4 N. = Force parallel to the plane.
Fv = 529*cos32 = 448.6 N. = Force perpendicular to the plane.
Fn = Fp - Fk
Fn = 280.4 - 0.310*448.6 = 141.3 N. =
Net force.
a = Fn/m = 141.3 / 54 = 2.62 m/s^2.
Wt of the block.
Fb = 529N @ 32 Deg. = Force of block.
Fp = 529*sin32 = 280.4 N. = Force parallel to the plane.
Fv = 529*cos32 = 448.6 N. = Force perpendicular to the plane.
Fn = Fp - Fk
Fn = 280.4 - 0.310*448.6 = 141.3 N. =
Net force.
a = Fn/m = 141.3 / 54 = 2.62 m/s^2.
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