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New answer posted
a month agoContributor-Level 10
For the combined system of mass M and m, the acceleration under an applied force F is:
a = F / (M + m)
The static friction force (f_s) on the top block (m) provides its acceleration:
f_s = MA = m * [F / (M + m)] = mF / (M + m)
For the top block not to slip, the required static friction must be less than or equal to the maximum possible static friction (μmg):
f_s ≤ μmg
mF / (M + m) ≤ μmg
F ≤ μ (M + m)g
Using the values implied in the solution:
F ≤ 21 N
New answer posted
a month agoContributor-Level 10
N = mg - F_L
f_s = mv²/R ≤ μsN = μs (mg - F_L)
F_L = m (v²/μsR - g)
New answer posted
a month agoContributor-Level 10
Mg + MkV² = MA = -mv (dV/dx)
Vdv = (−) (g + kV²)dx
∫? (Vdv)/ (g + kv²) = ∫? - dx
[ln (g + kV²)/2k]? = -x
ln (g/ (g + ku²) = −2kx
x = (1/2k)ln (1 + ku²/g)
New answer posted
a month agoContributor-Level 10
mdv? /dt = kv? (1) and mdv? /dt = kv?
(2)/ (1) ⇒ dv? /dv? = v? /v?
v? dv? = v? dv?
v? ² = v? ² + C
v? ² – v? ² = C = Constant
Now, v? * a? = (v? î + v? ) * (k/m) (v? î + v? )
= (k/m) [v? ²k? – v? ²k? ] = (k/m) (v? ² – v? ²)k? = Constant.
New question posted
a month agoNew answer posted
a month agoContributor-Level 10
σ4πr² + σ4πR² = Q ⇒ σ = Q/ (4π (R²+r²)
V_c = kq? /r + kq? /R = k (σ4πr²)/r + k (σ4πR²)/R = kσ4π (r+R)
= K (Q/ (R²+r²) (R+r)
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