<p>The equations used are:
1)ΣF=ma
2)fk = μkN
3)fs ≤ μsN
4)Fs = - kx</p>
<p>Here are the questions:
1. If ΣF is 10N and a=1 m/s^2, find m using the first equation
2. Given ΣF = fk, m=250 kg, μk=0.2, and N = 10m, find a.
3. ΣF=T-10m, but a=0 m/s^2. Use the first equation to find m in terms of T.
4. Given the following values, determine if the third equation is valid. ΣF =fs, m=90kg, and a=2 m/s^2. Also, μs=0.1 and N=5N.
5. Use the 4th equation to solve for F, if k=900 N/m and x=0.15m
6. Use this equation ( vf=v0+at) and (ΣF=ma) find ΣF when m=12kg, vo= 15m/s,vf=5/ms and x=0.15m</p>
<p>1) F = ma, 10 = m(1) → m = 10kg
2) Not really enough detail but I’ll just assume that it’s not any slope.
f = μmg = -0.2<em>250</em>9.8 = -490N
f = ma → f/m = a → -490/250 = a
a = -1.96 m/s²
3) When a = 0, ΣF = 0 because F = ma, and F = m(0) = 0, so F = 0
T-10m = 0
T = 10m
4) It’s not really an equation, but okay, and I’m probably misunderstanding this question because it lacks detail as to which is which…
fs ≤ μsN
fs = ma
ma = μsN
90<em>2 = 0.1</em>5
180 is not smaller than or equal to 0.5? I think?
Generally, it’s ■■ ≤ fs, but technically
fs ≤ μsN SHOULD be valid, but it’s probably more like fs = μsN, I’m probably just confused with which numbers to use.
5) F = -kx
F = -900<em>0.15
= -135N
6) We can’t calculate acceleration because we don’t have the time over which the speed changed from vo to vf.
Because a = (vf-vi)/t → We can’t calculate a, because we don’t know what t is.
Also, if x was actually the displacement Δx, then we could get somewhere, because
vf² = vo² + 2aΔx
(vf²-vi²)/(2Δx) = a
F = ma
= m(vf²-vi²)/(2Δx)
= 12(25 - 225)/(2</em>0.15)
F = -8000N</p>