Answer:
Explanation:
Let the bigger crate be in touch with the ground which is friction less. In the first case both m₁ and m₂ will move with common acceleration because m₁ is not sliding over m₂.
1 ) Common acceleration a = force / total mass
= 234 / ( 25 +91 )
= 2.017 m s⁻².
2 ) Force on m₁ accelerating it , which is nothing but friction force on it by m₂
= mass x acceleration
= 25 x 2.017
= 50.425 N
The same force will be applied by m₁ on m₂ as friction force which will act in opposite direction.
3 ) Maximum friction force that is possible between m₁ and m₂
= μ_s m₁g
= .79 x 25 x 9.8
= 193.55 N
Acceleration of m₁
= 193 .55 / 25
= 7.742 m s⁻²
This is the common acceleration in case of maximum tension required
So tension in rope
= ( 25 +91 ) x 7.742
= 898 N
4 ) In case of upper crate sliding on m₂ , maximum friction force on m₁
= μ_k m₁g
= .62 x 25 x 9.8
= 151.9 N
Acceleration of m₁
= 151.9 / 25
= 6.076 m s⁻².
I believe it’s the scientific revolution
Hope this helps :)
Answer:
The work done is 205 kJ.
Explanation:
Hi there!
Work can be calculated using the following equation:
W = F · Δx
Where:
W = work
F = applied force
Δx = displacement
In this case, the force varies with the position, so we can divide the traveled distance in very small parts and calculate the work done over each part of the trajectory. Then, we have to sum all the works and we will obtain the work done from the initial position (xi) to the final position (xf). This is the same as saying:
W = ∫ F · dx
F = 3.6 N/m³ · x³ - 76 N
W = ∫ (3.6 x³ - 76)dx
W = 0.9 x⁴ - 76x
Evaluating from xi to xf:
W = 0.9 N/m³ (21.9 m)⁴ - 76 N · 21.9 m - 0.9 N/m³(5.41 m)⁴ + 76 N · 5.41 m
W = 205 kJ
R=U^2/P=120*120/40=360 ohm
P2=U2^2/R=132*132/360=48.4 w
power increase ratio (48.4-40)/40=21%
