Answer:
Magnitude of the Frictional force = (mv₀²)/2x₁
Explanation:
For the frictional force to stop the box, it has to produce the deceleration of the box; thereby being the opposing force to the box's motion.
According to Newton's first law of motion
Frictional force = (mass of the box) × (deceleration experienced by the box)
Let the mass of the box be m
Then,
Frictional force = ma
Then we can obtain the deceleration using the equations of motion
v² = u² + 2ax
u = Initial velocity = v₀ m/s
v = Final velocity = 0 m/s (since the box comes to rest at the end)
x = horizontal distance covered = (x₁ - x₀) = x₁ (since x₀ = 0)
a = ?
v² = u² + 2ax
0 = (v₀)² + 2ax₁
2ax₁ = - v₀²
a = - (v₀²)/(2x₁) (minus sign, because it's a deceleration)
Magnitude of the Frictional force = ma = (mv₀²)/2x₁
Heyyyy I wish I could help but I don’t know
Answer:
Assume two identical cans filled with two types of soup having same mass are rolling down on an inclined plane in same conditions. In terms of inertia different types of soup will indicate different viscosity. The higher viscosity fillings indicates more part of the soup mass is rotating together with the can’s body. This means that for the can with lower viscosity soup has a lower moment of inertia and the can with higher viscosity has higher moment of inertia while the same gravity makes them to roll.
incline angle = θ ; can's mass = m ; Radius of the can's = R , Angular acceleration for Can 1 = α1 ; Angular acceleration for Can 2 = α2
T1 = Inertia of Can with high viscosity soup
T2 = Inertia of Can with low viscosity soup
M1 rolling moment of Can 1
M2 rolling moment of Can 2
equation is given by
T1*α1 = M1 - (a)
T2*α2 = M2 - (b)
M1 = M2 = m*g*R*sin(θ). (c)
as assumed T1 > T2
from the three equation (a), (b) & (c)
the α2 > α1
Angular acceleration of Can 2 is higher than Can 1. Already stated that Can 1 has more viscous soup as compared to Can 2.
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b. alpha
Answer:
thanks for the points man
