(a) What is the coefficient of kinetic friction between the crate and the floor?

Question

(a) What is the coefficient of kinetic friction between the crate and the floor?
?k =________


(b) If the 320-N force is instead pulling the block at an angle of 20.0

Explanation / Answer

Let W be the weight of the crate. and R the frictional force (a) You can separate the force F into a horizontal and vertical component: F_h = F·cos(20°) F_v = -F·sin(20°) The change of momentum in horizontal direction is equal to the sum of the forces acting in horizontal direction: dp/dt = F_h - R m·a = F·cos(20°) - µ·N N is the normal force exerted by the plane, which is equal to the sum of the vertical forces acting downwards on the crate: N = W + F·sin(20°) => m·a = F·cos(20°) - µ·(W + F·sin(20°)) Because the crate moves at constant velocity, the acceleration is zero. Hence: F·cos(20°) - µ·(W + F·sin(20°)) = 0 µ = F·cos(20°) / (W + F·sin(20°)) = 300N · cos(20°) / (1000N + 300N · sin(20°)) = 0.2557 (b) In this case F acts partially upward: F_h = F·cos(20°) F_v = F·sin(20°) This changes the normal force to: N = W - F·sin(20°) The overall momentum balance becomes to: m·a = F·cos(20°) - µ·(W - F·sin(20°)) a = [ F·cos(20°) - µ·(W - F·sin(20°)) ] / m The mass of the crate is given by W = m·g m = W /g Therefore the acceleration of the crate is: a = g · [ (F/W)·cos(20°) - µ·(1 - (F/W)·sin(20°)) ] = 9.81m/s² · [ 0.3 · cos(20°) - µ·(1 - 0.3·sin(20°)) ] = 0.51m/s²

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