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A 1,160-N crate is being pushed across a level floo t speed by a force F of 270

ID: 1884970 • Letter: A

Question

A 1,160-N crate is being pushed across a level floo t speed by a force F of 270 N at an angle of 20.0° below the 20.0 20.0° (a) What is the coefficient of kinetic friction between the crate and the floor? (Enter your answer to at least three decimal places.) (b) If the 270-N force is instead pulling the block at an angle of 20.0° above the horizontal, as shown in the figure b, what will be the acceleration of the crate? Assume that the coefficient of friction is the same as that found in part (a) Read It Show My Work (Optional)

Explanation / Answer

here,

weight of crate , W = 1160 N

force , F= 270 N

theta = 20 degree

a)

let the coefficient of kinetic friction be uk

as it is moving at constant speed

the net force is zero

F * cos(theta) = uk * ( W + F * sin(theta))

270 * cos(20) = uk * ( 1160 + 270 * sin(20))

solving for uk

uk = 0.20

b)

the accelration of the system , a = net force /effective mass

a = (F * cos(theta) - uk * ( W - F * sin(theta)) /(W/g)

a = ( 270 * cos(20) - 0.2025 * ( 1160 - 270 * sin(20))) /( 1160/9.81)

a = 0.32 m/s^2