Question

This crate has a weight of 200 pounds. The kinetic coefficient between it and the floor is 0.45. If a constant force P of 120 pounds is applied as shown, how far will the crate move in 5 seconds, if it is initially at rest? (Give your answer in units of feet). 

Answer 1

Answer 2

Giveno M W = 200 lb P= 120 lb. A) Cena Fm عي 9 M=0.45. t = 5 see isitial velocity ,u=0. (aritially rest) Solution Applying the equations of dynamis. EF=ma 0. IP-Fm = ma Ev=o. NR=W 200 lb limiting friction fm = MNR 0.45 x 200 fm = 90 lb.

Answer 3

Answer 4

And ay initial velocity, u=0 N. a. P=120 lb o=ff} agoon=N Me=o.us fr 200lb. row = and then I would solue ti Efre » P- fx in CN' 8 m' your units. suurt convert it into P=533-787 N m block 90.718 box g= 9.81 m/s2 6. N= mg -889,94N 60 Efr zy 533.787 = 90.7 18 a brunt 533-787-400.474 =90-718 a a = 133.31 90.71 = 1.47 m/s²

and t S Дес. yo at . 3 Displacement of д- у - ав? Т- in thai sexy as Sek• 8 авч 25 teams ملا **, Үхэх bt.o9 25 Я д 09 around feet

Answer 5

The crate moves in 5 seconds in 1.875 feet.

Giveno M W = 200 lb P= 120 lb. A) Cena Fm عي 9 M=0.45. t = 5 see isitial velocity ,u=0. (aritially rest) Solution Applying the equations of dynamis. EF=ma 0. IP-Fm = ma Ev=o. NR=W 200 lb limiting friction fm = MNR 0.45 x 200 fm = 90 lb.

substitute * is 0. ma 120 - 90 30 lb = 200 lb xa 0.15 f4 /3 =a. using the equation, 8 = ut + Çat? = 0 + 1 x 0,155 8 = 1.875 Ft.