Expert Answer Step 1 (d) The weight of the crate is W=mg =(40 kg) (9. 81 "/ ) =392. 4 N 392 N Step 2 (e) The crate is moving in horizontal direction only. So, the sum of vertical components Normal and the weight of the crate cancels out. From the free body diagram, it is obvious that the only horizontal force acting on the crate is the frictional force. Therefore, the net force on the crate is the frictional force, Fnet=-F, =-313. 92 N Negative sign because of the direction of the force Step 3 (f) From Newton's law the deceleration of the crate is a=4 313.92 N 40 kg =7. 848 "/ As per guide lines first 3 sub parts are solved. kindly post remaining as a new post.

Expert Answer Step 1 (d) The weight of the crate is W=mg =(40 kg) (9. 81 "/ ) =392. 4 N 392 N Step 2 (e) The crate is moving in horizontal direction only. So, the sum of vertical components Normal and the weight of the crate cancels out. From the free body diagram, it is obvious that the only horizontal force acting on the crate is the frictional force. Therefore, the net force on the crate is the frictional force, Fnet=-F, =-313. 92 N Negative sign because of the direction of the force Step 3 (f) From Newton's law the deceleration of the crate is a=4 313.92 N 40 kg =7. 848 "/ As per guide lines first 3 sub parts are solved. kindly post remaining as a new post.