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ivolga24 [154]
3 years ago
6

A 53 kg crate is at rest on a level floor, and the coefficient of kinetic friction is 0.36. The acceleration of gravity is 9.8 m

/s 2 . If the crate is pushed horizontally with a force of 372.098 N, how far does it move in 4.94 s? Answer in units of m.
Physics
1 answer:
tino4ka555 [31]3 years ago
6 0

Answer:

42.6 m

Explanation:

mass of crate m = 53 kg

coefficient of kinetic friction, μ = 0.36

acceleration due to gravity, g = 9.8 m/s^2

Force, F = 372.098 N

Net force, f = F - friction force

f = 372.098 - μ m x g = 372.098 - 0.36 x 53 x 9.8

f = 185.114 N

acceleration, a = f / m = 185.114 / 53 = 3.49 m/s^2

initial velocity, u = 0

time, t = 4.94 s

s = ut + 1/2 at^2

s = 0 + 1/2 x 3.49 x 4.94 x 4.94

s = 42.6 m

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Answer:

a)   x₁ = 290.50 feet ,  x₂ = 169.74 feet , b)  v_max= 41 mph

Explanation:

For this exercise we will work in two parts, the first with Newton's second law to find the acceleration of vehicles

X Axis          fr = m a

Y Axis          N-W = 0

                    N = W = mg

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                  fr = μ N

We replace

                 μ mg = ma

                 a = μ g

                 g = 32 feet / s²

Let's calculate the acceleration for each coefficient and friction

μ              a (feet / s2)

0.599       19.168

0.536       17,152

0.480       15.360

0.350        11.200

These are the acceleration values, for the maximum distance we use the minimum acceleration (a₁ = 11,200 feet / s²) and for the minimum braking distance we use the maximum acceleration (x₂ = 19,168 feet / s²)

                 v² = v₀² - 2 a x

When the speed stops it is zero

                 x₁ = v₀² / 2 a₁

                         

Let's reduce speed

            v₀ = 55mph (5280 foot / 1 mile) (1h / 3600s) = 80,667 feet / s²

Let's calculate the maximum braking distance

            x₁ = 80.667² / (2 11.2)

            x₁ = 290.50 feet

The minimum braking distance

            x₂ = 80.667² / (2 19.168)

            x₂ = 169.74 feet

b) maximum speed to stop at distance x = 155 feet

            0 = v₀² - 2 a x

            v₀ = √2 a x

We calculate the speed for the two accelerations

             v₀₁ = √ (2 11.2 155)

             v₀₁ = 58.92 feet / s

       

             v₀₂ = √ (2 19.168 155)

             v₀₂ = 77.08 feet / s

To stop at the distance limit in the worst case the maximum speed must be 58.92 feet / s = 40.85 mph = 41 mph

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