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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

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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Salsk061 [2.6K]

1) 1.2 m/s

First of all, we need to find the angular velocity of the blade at time t = 0.200 s. This is given by

$\omega_f = \omega_i + \alpha t$

where

$\omega_i = 0.300 rev/s$ is the initial angular velocity

$\alpha = 0.895 rev/s^2$ is the angular acceleration

Substituting t = 0.200 s, we find

$\omega_f = 0.300 + (0.895)(0.200)=0.479 rev/s$

Let's now convert it into rad/s:

$\omega_f = 2\pi \cdot 0.479 rev/s=3.01 rad/s$

The distance of a point on the tip of the blade is equal to the radius of the blade, so half the diameter:

$r=\frac{0.800}{2}=0.400 m$

And so now we can find the tangential speed at t = 0.200 s:

$v=\omega_f r =(3.01)(0.400)=1.2 m/s$

2) $2.25 m/s^2$

The tangential acceleration of a point rotating at a distance r from the centre of the circle is

$a_t = \alpha r$

where $\alpha$ is the angular acceleration.

First of all, we need to convert the angular acceleration into rad/s^2:

$\alpha = 0.895 rev/s^ \cdot 2 \pi =5.62 rad/s^2$

A point on the tip of the blade has a distance of

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From the centre; so, the tangential acceleration is

$a_t = (5.62)(0.400)=2.25 m/s^2$

3) $3.6 m/s^2$

The centripetal acceleration is given by

$a=\frac{v^2}{r}$

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v is the tangential speed

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We already calculate the tangential speed at point a):

v = 1.2 m/s

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7 0

3 years ago

I need help with 23 please.

Ne4ueva [31]

The answer is c.

hope this helps! :)

8 0

3 years ago

Anon25 [30]

I thinks its period because they are on the last column to the right

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3 years ago

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. A person weighing 750 N gets on an elevator.

Kobotan [32]

F = 750 N (Force)

d = 10 m (displacement )

t = 25 s (time)

L = ? (Mechanical work ) = (Energy)

P = ? (Power)

Solve:

L = F × d = 750 × 10 = 7500 Joules

P = L / t = 7500 / 25 = 300 Watts

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2 years ago

When reaching a boundary between two media (1 and 2), an incident ray is partially reflected and partially refracted. The ray is

lukranit [14]

Answer:

The angle of incidence when the reflected ray is perpendicular to the incident ray = 45°

Explanation:

According to Snell's Law,

n₁ sin θ₁ = n₂ sin θ₂

When the angle between the incident ray and reflected ray is 90°, the angle of incidence is θ₁ and the angle of reflection, θ₂ = 90° - θ₁ and the index of refraction in the Snell's Law for both media would be the same, n₁ = n₂ = n

n sin θ₁ = n sin (90° - θ₁)

Note that from trigonometric relations,

Sin (90° - θ₁) = cos θ₁

n sin θ₁ = n cos θ₁

(sin θ₁)/(cos θ₁) = 1

tan θ₁ = 1

θ₁ = arctan 1 = 45°

Hope this Helps!!!

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