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Crazy boy [7]
3 years ago
9

What is the net force acting on a 52 kg object that has a velocity of 8.0 m/s and is moving in a circle of radius 1.6 m? a. 4000

N b. 20880N c. 2500N d. 3500N
Physics
1 answer:
jenyasd209 [6]3 years ago
7 0
Σf = m a
Σf = m v^2 / r
Σf = 52 8^2 / 1.6
Σf = 2080 N
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Harrizon [31]

Answer:

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b) Acceleration of the toboggan is going downhill = 4.76 m/s²

Explanation:

Resolving the normal reaction into vertical and horizontal component

Nₓ = mg sin θ

Nᵧ = mg cos θ

The frictional force on the toboggan = Fr = μmg cos θ (note that μ = μ(k) = the coefficient of kinetic friction)

Doing a Force balance on the x component taken in the axis parallel to the inclined plane.

The net force that has to accelerate the toboggan has to match the frictional force acting downwards of the plane (In the opposite direction to motion) and x-component of the Normal reaction

ma = Nₓ + Fr

ma = (mg sin θ) + (μmg cos θ)

a = g[(sin 44°) + (0.29 cos 44°)

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b) While coming downhill,

The frictional force is acting uphill now (In the direction opposite to motion),

The force balance is

ma = (mg sin θ) - (μmg cos θ)

a = g[(sin 44°) - (0.29 cos 44°)]

a = 9.8(0.6947 - 0.2086)

a = 4.76 m/s²

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geniusboy [140]

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AleksAgata [21]

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