My answer for this question is d as every car has the same result for the force of friction since the normal and coefficient of static friction is the same. I cannot find an answer online so can anyone help verify this? Thank you.
Finding x by force formula
- only force acting is gravity
ma/-k = x
(0.2)(-9.8)/185 = x
0.010594594 = x
Finding x by wd formula
WD_ spring = (1/2)kx^2
F x = (1/2)kx^2
2(mg)/k = x
[2(0.2)(-9.8)]/ 185 = x
0.021189189 = x
how come the work done and force formulas produce different values for x...
Ok thank you, i edited the forum. I guess you can't open pictures on this website, but i calculated d with Vf^2=Vi^2 +(2)(a)(d)
0^2 = 15^2 + (2)(-9.8)d
d = 11.4m
(I realize i probably shouldn't use "x" and "d" interchangibly)
I am just confused on how to find the normal force/ FN of the first object. My classmates are saying Fgy is the exact same as Fn but I don’t get why
Fgy= Fg sin theta
Fgy= (20)(9.81) (sin35)
Fgy= 112.5
Fgy = FN
1. break down 430N [Up 35* L] into components
430 cos 35 = 352.2N [L]
430 sin 35 = 246.6N
2.
ΣFx= 352.2N [L]
ΣFy= 246.6N + 280N - 430N
ΣFy = 86.6
3.
Pythagorean theorem with the two will give you a magnitude of 362.7N
Then using tan you can find the angle of [L 76 U]
This method was...
Sir, if you looked at the picture I sent you:
- You would've saw my vectors; I added two opposite vectors meaning I changed the 2.78W into -2.78E . I did that to simply save space and time.
- I also did (6.25)(cos40) = 4.788 and I am not sure if you're telling me not to round from 4.787777769...