Question

b) Determine the acceleration of the box Questions (2 marks) State Newton's First Law of Motion oure 2 shows a box puted by a force F on a rough horizontal surface The GOR es with constant velocity with presence of 5 N frictional force Figure 2 Ilustrate a labeled free-body diagram of the box. (4 marks) Determine the force, F. (3 marks) Compute the reaction force and the mass of the object if the coefficient of friction is 0.15. (6 marks)

Can you help me answer all these questions.thank you so much.your help is highly appreciated

Answer 1

Question (3) (a) Newton's first law state an object will remain art rest or in uniform motion in straight line unless acted upon by a force an External force. (6) 4 (1) free body diagron of box V = cont (1) F= ? As Velocity = constant TESH Jing FNet in Horizental direction so эЕ Flosso=f Fa 5 Coszoceszo 0 f lesso o nnector F=5+32 Newton / f = Sh 4 .989 f= t - ( Esthro rh F N+Fs1n2o = mg e. pl. frict pentical to 20 N=mg-fsinzo 8Nsf Vong as frechen force f= UN 5=0:15 (mg - FS1920) where F=3.32 H & sa 5=0.15(m (9.8)-5-32 x 0.34) sin20=0.34 5 = m (9.8) - 1.8 33.33 +1.8 m 1m = 3.58 kg 9.8 O'15 N = ing - FS1n2o = (3.58x9.8) - 5.32*0.34 = 35.084 - 1.8 IN = 33.28 Newton/

Question 4 (9) Normal Reaction & An Electromagnetic force which act beth two surface in contact due to interaction bets molecules of two surfaces this orce always act loret perpendicular to surface in contact e towards body aco (1) free body diagrams const osulsoon So. J 40o casso Iso Falsh (1) As 950s Vlucon K wt of box fret parallel to plakee o x = uuosinsotf x = 400 sinsootis Questions x=40oX0-76+15 x= 319 N Hes free body Dragram aft I ya coth so asuis bh SON 9 = 549.8 0=49N (1) H= ug cosso de Mich dx = 0.25 x ug cosso = 0.25x 49 x 0664 I fx = 7.8487