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olga55 [171]
3 years ago
13

150-N box is being pulled horizontally in a wagon accelerating uniformly at 3.00 m/s2. The box does not move relative to the wag

on, the coefficient of static friction between the box and the wagon's surface is 0.600, and the coefficient of kinetic friction is 0.400. The friction force on this box is closest to_________
Physics
1 answer:
mafiozo [28]3 years ago
3 0

Answer:

the friction force on this box is closest to 45.9 N.

Explanation:

given data

Weight of the box W = 150 N

accelerating uniformly = 3.00 m/s²

coefficient of kinetic friction = 0.400

coefficient of static friction = 0.600

solution

we know box does not move relative to the wagon

we get here friction force  that is express as

friction force = mass × acceleration ..............1

here mass = weight ÷ g

mass = \frac{150}{9.8} = 15.3 kg

put value in equation 1 we get

friction force = 15.3 kg × 3

friction force =  45.9 N

So, the friction force on this box is closest to 45.9 N.

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A 3.0 kg box is sliding along a frictionless horizontal surface with a speed of 1.8 m/s when it encounters a spring. (a) Determi
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Answer:

a) k = 2231.40 N/m

b) v = 0.491 m/s

Explanation:

Let k be the spring force constant , x be the compression displacement of the spring and v be the speed of the box.

when the box encounters the spring, all the energy of the box is kinetic energy:

the energy relationship between the box and the spring is given by:

1/2(m)×(v^2) = 1/2(k)×(x^2)

    (m)×(v^2) = (k)×(x^2)

a) (m)×(v^2) = (k)×(x^2)

                 k = [(m)×(v^2)]/(x^2)

                 k = [(3)×((1.8)^2)]/((6.6×10^-2)^2)

                 k = 2231.40 N/m

Therefore, the force spring constant is 2231.40 N/m

b) (m)×(v^2) = (k)×(x^2)

             v^2 = [(k)(x^2)]/m

                 v =  \sqrt{ [(k)(x^2)]/m}

                 v = \sqrt{ [(2231.40)((1.8×10^-2)^2)]/(3)}

                    = 0.491 m/s

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