Asked by Anonymous
A box of books weighing 300N is shoved the floor of an apartment by a force of 400N exerted downward at an angle of 35.2 degrees below the horizontal. If the coefficient of kinetic friction between box and floor is .570, how long does it take to move the box 4.00 m, starting from rest?
Work:
(.570)300N=m(9.8m/s^2)
m=14.4 kg
400N=(30.6kg)a
a=27.8m/s^2
4m=(1/2)(27.8m/s^2)(sin 35.2)t^2
t=.71 sec
However, the answer is 3.16s.
Work:
(.570)300N=m(9.8m/s^2)
m=14.4 kg
400N=(30.6kg)a
a=27.8m/s^2
4m=(1/2)(27.8m/s^2)(sin 35.2)t^2
t=.71 sec
However, the answer is 3.16s.
Answers
Answered by
bobpursley
No. The shoving force has a horizontal component, and a vertical component. The vertical component adds to the weight to make a greater fricton force.
vertical component=300*Sin35.2
horizontal component= 300 Cos35.2
Friction= mu(mg + vertical component)
net horizontal force=ma
300cos35.2-friction= ma
solve for a.
Then, put it in the distance equation.
vertical component=300*Sin35.2
horizontal component= 300 Cos35.2
Friction= mu(mg + vertical component)
net horizontal force=ma
300cos35.2-friction= ma
solve for a.
Then, put it in the distance equation.
Answered by
franca
how long does it take to move the box 4.00m starting from rest?
Answered by
Anonymous
bob ur wrong
Answered by
bob
1.13 seconds
Answered by
Vivek
300 cos 35.2 - Fk = ma
fk = 300 cos 35.2 - ma
Fn = mg + 400 sintheta = 300 + 400 sin 35.2
Fk = Fn*uk
uk = 0.57
300 cos 35.2 - ma = 0.57 [ 300 + 400sin 35.2]
-ma = (0.57)[300+400sin35.2]- [300 cos 35.2]
a = all that crap on the right side/ (300N/0.9)
a = 0.8 m/s2
Sx = VoT+ 1/2 at^2
4= 0+ 1/2 (0.8) t^2
t2 = 10
t = 3.16 s
fk = 300 cos 35.2 - ma
Fn = mg + 400 sintheta = 300 + 400 sin 35.2
Fk = Fn*uk
uk = 0.57
300 cos 35.2 - ma = 0.57 [ 300 + 400sin 35.2]
-ma = (0.57)[300+400sin35.2]- [300 cos 35.2]
a = all that crap on the right side/ (300N/0.9)
a = 0.8 m/s2
Sx = VoT+ 1/2 at^2
4= 0+ 1/2 (0.8) t^2
t2 = 10
t = 3.16 s