A 100-kg box is towed to move horizontally from rest by a constant force P=200 N. The kinetic friction is μk =0.1. The angle of the force P is θ=30° with respect to the horizontal direction. The acceleration due to gravity is g=9.81 m/s2. (2) Select the motion of equation along the vertical direction, where F is the friction force, N is the normal force from the ground, and a is the acceleration.___________ A.Psin30°-mg+N=ma B.Psin30°-mg+N=0 C.Pcos30°-mg+N=ma D.Pcos30°-mg+N=0 A 100-kg box is towed to move horizontally from rest by a constant force P-200 N. Thekinetic friction is μ =0.1. The angle of the force P is 0-30° with respect to the horizontaldirection. The acceleration due to gravity is g=9.81 m/s²(2) Select the motion of equation along the vertical direction, where F is the friction force,N is the normal force from the ground, and a is the acceleration.A. Psin30°-mg+N=maB. Psin30°-mg+N=0OC. Pcos30°-mg+N=maO D. Pcos30°-mg+N=0

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Answered: A 100-kg box is towed to move…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

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A 100-kg box is towed to move horizontally from rest by a constant force P=200 N. The kinetic friction is μ_k =0.1. The angle of the force P is θ=30° with respect to the horizontal direction. The acceleration due to gravity is g=9.81 m/s².

(2) Select the motion of equation along the vertical direction, where F is the friction force, N is the normal force from the ground, and a is the acceleration.___________

	A.	Psin30°-mg+N=ma
	B.	Psin30°-mg+N=0
	C.	Pcos30°-mg+N=ma
	D.	Pcos30°-mg+N=0

A 100-kg box is towed to move horizontally from rest by a constant force P-200 N. The
kinetic friction is μ =0.1. The angle of the force P is 0-30° with respect to the horizontal
direction. The acceleration due to gravity is g=9.81 m/s²
(2) Select the motion of equation along the vertical direction, where F is the friction force,
N is the normal force from the ground, and a is the acceleration.
A. Psin30°-mg+N=ma
B. Psin30°-mg+N=0
OC. Pcos30°-mg+N=ma
O D. Pcos30°-mg+N=0

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