Question

12 m/s When the 9 kg box reaches point it has a speed Neglect friction and the sure of the box. (Figure 1) Part A Determine how high the box reaches up the surface before it comes to a stop Express your answer to three significant figures and include the appropriate units. HAR O ? Value Units Figure Submit Request Answ Part B What is the resultant normal force on the surface at this point? Express your answer to three significant figures and include the appropriate units. Á + Om N = Value Units

Answer 1

sos Given mass, m= 9kg initial velocity, VA = Rm/s equation of path, dat dy=3 Assumption & No friction force am lub >N 80-90 12m/s 190 gm (A) for finding heighest we can use energy let assume it height at which it stops conservation relation. stops at point B. So, I mv & mgha = 1 m ve tamgh VA = velocity at A B = velocity at B HA= height at A ho = height at B - tm (12) ² + 0 = 0 + ng ha 144 = ghB g = 9.8 m/s2 2 hR= 144 ~ 2x9.8 7.3469 metres 7.347 metres ho

a axis. (B) we will find angle o made by tangent line at B with let that angle be & as shown in fig. x + y =3 differentiate wist & - 7 2 Y2 + 1 z dy = 0 & Y dy = - X Y Z » e V --VIE , wing - du -9.3634 tandu-9.3634 p=tant (-9. 3634) $ = -83.9039° A 83.9039 V-83.90390 903 90 here are sign with angle Shows its is measured clockwis from x axis. so 83-9039° = 180-0 & from figure. => 0 = 96.096° Angle force of gravity makes with with tangent is 0-90° - 6.096°

Applying Newton's and how along Normal direction at B.. mg sin (0-90°) = N N = 9x9.8 * sin(6.096°) 9.3664 Newtons