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kari74 [83]
3 years ago
12

a box having a mass of 50 kg is dragged across a horizontal floor by means of a rope tied on the front of it. the coefficient of

friction between the box and the floor is .3. if the angle between the rope and the floor is 30 degrees, what force must be exerted on the rope to move the box with an acceleration of 2.5 m/second squared
Physics
1 answer:
Rina8888 [55]3 years ago
3 0

Answer: 148.263 N

Explanation:

If we draw a free body diagram of the describes situation we will have the following:

Net force in x:

-F_{fr}+Fcos\theta=0 (1)

Where:

F_{fr}=\mu_{k}N is the friction force

F is the force exerted on the rope to move the box

\theta=30\° is the angle

Net force in y:

N+Fsin\theta-W=0 (2)

Where:

N is the Normal force

W=mg is the force due gravity on the box (its weight), being m=50 kg the mass of the box and g=9.8 m/s^{2} is the acceleration due gravity

Rewritting (2):

N+m a sin\theta-mg=0 (3)

Where a=2.5 m/s^{2}

N=m(g-a sin\theta) (4)

N=50 kg(9.8 m/s^{2}-(2.5 m/s^{2}) sin30\°) (5)

N=428 N (6) This is the value of the normal force

Rewritting (1):

-\mu_{k}N+Fcos\theta=0 (6)

Finding F:

F=\frac{\mu_{k}N}{cos 30\°} (7)

F=\frac{(0.3)(428 N)}{cos 30\°}

Finally:

F=148.263 N

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3 years ago
A solid sphere, a solid disk, and a thin hoop are all released from rest at the top of the incline (h0 = 20.0 cm).
Ede4ka [16]

Answer:

a. The object with the smallest rotational inertia, the thin hoop

b. The object with the smallest rotational inertia, the thin hoop

c.  The rotational speed of the sphere is 55.8 rad/s and Its translational speed is 1.67 m/s

Explanation:

a. Without doing any calculations, decide which object would be spinning the fastest when it gets to the bottom. Explain.

Since the thin has the smallest rotational inertia. This is because, since kinetic energy of a rotating object K = 1/2Iω² where I = rotational inertia and ω = angular speed.

ω = √2K/I

ω ∝ 1/√I

since their kinetic energy is the same, so, the thin hoop which has the smallest rotational inertia spins fastest at the bottom.

b. Again, without doing any calculations, decide which object would get to the bottom first.

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= 0.03 m × 55.8 rad/s

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