A 1,380-N crate is being pushed across a level floor at a constant speed by a force F of 320 N at an angle of 20.0° below the horizontal, as shown in the figure a below.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 320-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).m/s2 The image consists of two diagrams labeled (a) and (b), each depicting a crate on a surface with a force vector \(\vec{F}\) applied at an angle. Diagram (a):- The crate is shown with a force \(\vec{F}\) acting on it.- The vector \(\vec{F}\) is directed downward at an angle of \(20.0^\circ\) relative to the horizontal surface.Diagram (b):- Similarly, a crate is depicted with a force \(\vec{F}\) acting on it.- The vector \(\vec{F}\) is directed upward at an angle of \(20.0^\circ\) relative to the horizontal surface.Both scenarios illustrate how the direction of the applied force at the same angle affects the interaction between the crate and the surface.

Name: A 1,380-N crate is being pushed across a level floor at a constant spe
Uploaded: 2024-03-20T06:56:46.000Z
Channel: Bartleby
Description: A 1,380-N crate is being pushed across a level floor at a constant speed by a force F of 320 N at an angle of 20.0° below the horizontal, as shown in the figure a below.a) What is the coefficient of kinetic friction between the crate and the floor? (

Formula used:- Fk = μk.n .......(1) μk is the kinetic friction The force is separated in two parts which is horizontal and vertical. now the force = the product of mass and acceleration. ∑F=m×aacceleration is zero∑F=0 this implies ∑Fx and ∑Fy=0, where Fx and Fy are horizontal and vertical components. Sine is acting on vertical part and mg force is applied on it and cosine is applied on horizontal part. ∑Fx=0320cos200-fk=0fk=300.8 now, vertical element ∑Fy=0n-fsin200-mg=0n-320sin200-1380=0n=109.446+1380n=1489.45 N Using eqn 1 in formula Fk = μk.n rearranging for kinetic friction μk=Fknsubstituting value=300.81489.45=0.2 (b) ∑Fy=0n+fsin200-mg=0n+320sin200-1380=0n=1270.5 N now, ∑Fx=maFcos200-fk=ma320cos200-μkn=ma.........(2) weight = mass(gravity) mass = weightgravity substituting value m = 138010 putting in (2) for acceleration 320cos200-μkn=138010a rearranging for acceleration a=320cos200-0.2(1270.5)138010a=0.34 ms2 Answer:- (a) kinetic friction is 0.2 (b) acceleration is 0.34ms2