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Komok [63]
3 years ago
8

A horizontal 52-n force is needed to slide a 50-kg box across a flat surface at a constant velocity of 3.5 m/s. What is the coef

ficient of kinetic friction between the box and the floor?
Physics
1 answer:
3241004551 [841]3 years ago
6 0

Answer: 0.61

Explanation:

This is calculation based on friction.

Since the box rests on a flat surface, the force that exists between them is known as frictional force.

Since the friction is dynamic (velocity is not zero)

The frictional force = kinetic energy gained by the body.

Ff = 1/2mv^2

coefficient of kinetic friction × normal reaction = 1/2mv^2

Since normal reaction is equal to the weight(force acting along the vertical component)

Normal reaction= mg = 50 × 10 = 500N. Therefore,

coefficient of kinetic friction × 500 = 1/2×50×3.5^2

coefficient of kinetic friction = 50×3.5^2/1000

coefficient of kinetic friction= 0.61

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Read 2 more answers
A 221 g cart starts from rest and rolls in the right direction (positive) down an incline. The incline is at a height of 5 cm. A
Julli [10]

Answer:

1) p₀ = 0.219 kg m / s, p = 0, 2)  Δp = -0.219 kg m / s, 3) 100%

Explanation:

For the first part, which is speed just before the crash, we can use energy conservation

Initial. Highest point

            Em₀ = U = mg y

Final. Low point just before the crash

           Emf = K = ½ m v²

          Em₀ = Emf

          m g y = ½ m v²

           v = √ 2 g y

Let's calculate

           v = √ (2 9.8 0.05)

           v = 0.99 m / s

1) the moment before the crash is

           p₀ = m v

           p₀ = 0.221 0.99

           p₀ = 0.219 kg m / s

After the collision, the car's speed is zero, so its moment is zero.

           p = 0

2) change of momentum

           Δp = p - p₀

            Δp = 0- 0.219

            Δp = -0.219 kg m / s

3) the reason is

     Δp / p = 1

In percentage form it is 100%

3 0
4 years ago
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