Physics Laws of Motion
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New answer posted
a month agoContributor-Level 9
Apply conservation of momentum along y-axis, we can write
10v? - 10v? sin 30° = 0
⇒ v? = 20m/s
New answer posted
a month agoContributor-Level 10
Assuming the rope in the boy's hand is vertical and using a Free Body Diagram (FBD):
f? = T
R + T = 90 ⇒ R = 90 - T
For the piece of wood not to move, f? ≤ µR:
T ≤ 0.5 (90 - T) ⇒ T ≤ 30N
New answer posted
a month agoContributor-Level 10
The minimum force F? is calculated as:
F? = (μmg) / √* (1 + μ²)* = ( (1/√3) * 1 * 10 ) / √* (1 + (1/√3)²) = 5N
New answer posted
a month agoContributor-Level 10
For an elastic collision where C comes to rest, and the compression in the spring is maximum, the velocities of A and B are equal (v). Using the conservation of mechanical energy:
(1/2)mv? ² = 2 * (1/2)mv² + (1/2)kx²
This gives the maximum compression x as:
x = v? * √* (m / 2k)*
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.
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