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3 years ago

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A fully loaded, slow-moving freight elevator has a cab with a total mass of 1400 kg, which is required to travel upward 37 m in

3.6 min, starting and ending at rest. The elevator's counterweight has a mass of only 930 kg, so the elevator motor must help pull the cab upward. What average power is required of the force the motor exerts on the cab via the cable?

1 answer:

kogti [31]3 years ago

6 0

Answer:

789.8 W

Explanation:

mass of the cab = 1400 kg, the counter weight of the elevator = 930 kg

weight of the cab = 1400 × 9.81 where weight = mg and m is mass and g is acceleration due to gravity.

weight of the cab = 13734 N

counter weight of the elevator = 930 × 9.81 = 9123.3 N

the exerted force of the elevator = weight of the cab - counter weight of the elevator = 13734 - 9123.3 = 4610.7 N

Average power by the motor P = F × v = F × distance / time

where v is speed in m/s, and time is in seconds

P = 4610.7 × 37 / ( 3.6 × 60) = 789.80 W

where (3.6 × 60 ) is the time in seconds

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4. A ball is thrown with an initial speed vi at an angle θi with the horizontal. The

GalinKa [24]

Disclaimer: I ended up finding what's asked for in the reverse order (e)-(a).

At time $t$ , the horizontal position $x$ and vertical position $y$ of the ball are given respectively by

$x = v_i \cos(\theta_i) t$

$y = v_i \sin(\theta_i) t - \dfrac g2 t^2$

and the horizontal velocity $v_x$ and vertical velocity $v_y$ are

$v_x = v_i \cos(\theta_i)$

$v_y = v_i \sin(\theta_i) - gt$

The ball reaches its maximum height with $v_y=0$ . At this point, the ball has zero vertical velocity. This happens when

$v_i \sin(\theta_i) - gt = 0 \implies t = \dfrac{v_i \sin(\theta_i)}g$

which means

$y = \dfrac R6 = v_i \sin(\theta_i) \times \dfrac{v_i \sin(\theta_i)}g - \dfrac g2 \left(\dfrac{v_i \sin(\theta_i)}g\right)^2 \\\\ \implies R = \dfrac{6{v_i}^2 \sin^2(\theta_i)}g - \dfrac{3{v_i}^2 \sin^2(\theta_i)}g \\\\ \implies R = \dfrac{3{v_i}^2 \sin^2(\theta_i)}g$

At the same time, the ball will have traveled half its horizontal range, so

$x = \dfrac R2 = v_i \cos(\theta_i) \times \dfrac{v_i \sin(\theta_i)}g \\\\ \implies R = \dfrac{2{v_i}^2 \cos(\theta_i) \sin(\theta_i)}g$

Solve for $v_i$ and $\theta_i$ :

$\dfrac{3{v_i}^2 \sin^2(\theta_i)}g = \dfrac{2{v_i}^2 \cos(\theta_i) \sin(\theta_i)}g \\\\ \implies 3 \sin^2(\theta_i) = 2 \cos(\theta_i) \sin(\theta_i) \\\\ \sin(\theta_i) (3\sin(\theta_i) - 2 \cos(\theta_i)) = 0$

Since $0^\circ$ , we cannot have $\sin(\theta_i)=0$ , so we're left with (e)

$3 \sin(\theta_i) - 2\cos(\theta_i) = 0 \\\\ \implies 3 \sin(\theta_i) = 2\cos(\theta_i) \\\\ \implies \tan(\theta_i) = \dfrac23 \\\\ \implies \boxed{\theta_i = \tan^{-1}\left(\dfrac23\right) \approx 33.7^\circ}$

Now,

$\cos\left(\tan^{-1}\left(\dfrac23\right)\right) = \dfrac3{\sqrt{13}}$

$\sin\left(\tan^{-1}\left(\dfrac23\right)\right) = \dfrac2{\sqrt{13}}$

so it follows that (d)

$R = \dfrac{2{v_i}^2 \times\frac3{\sqrt{13}} \times \frac2{\sqrt{13}}}g \\\\ \implies {v_i}^2 = \dfrac{13Rg}{12} \\\\ \implies \boxed{v_i = \sqrt{\dfrac{13Rg}{12}}}$

Knowing the initial speed and angle, the initial vertical component of velocity is (c)

$v_y = \sqrt{\dfrac{13Rg}{12}} \sin\left(\tan^{-1}\left(\dfrac23\right)\right) \\\\ \implies v_y = \sqrt{\dfrac{13Rg}{12}} \times \dfrac2{\sqrt{13}} \\\\ \implies \boxed{v_y = \sqrt{\dfrac{Rg}3}}$

We mentioned earlier that the vertical velocity is zero at maximum height, so the speed of the ball is entirely determined by the horizontal component. (b)

$v_x = \sqrt{\dfrac{13Rg}{12}} \times \dfrac3{\sqrt{13}} \\\\ \implies v_x = \dfrac{\sqrt{3Rg}}{2}$

Then with $v_y=0$ , the ball's speed $v$ is

$v = \sqrt{{v_x}^2 + {v_y}^2} \\\\ \implies v = v_x \\\\ \implies \boxed{v = \dfrac{\sqrt{3Rg}}2}$

Finally, in the work leading up to part (e), we showed the time to maximum height is

$t = \dfrac{v_i \sin(\theta_i)}g$

but this is just half the total time the ball spends in the air. The total airtime is then

$2t = \dfrac{2 \times \sqrt{\frac{13Rg}{12}} \times \frac2{\sqrt{13}}}g \\\\ \implies 2t = 2\sqrt{\dfrac R{3g}}$

and the ball is in the air over the interval (a)

$\boxed{0 < t < 2\sqrt{\frac R{3g}}}$

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2 years ago

22. A race car accelerates from 0.0 m/s to 5 m/s with a displacement of 2.5 m. What is the

yanalaym [24]

Answer:

Option A

$5 m/s^{2}$

Explanation:

From fundamental equation of motion

$v^{2}=u^{2}+2as$ where v is the final velocity, u is the initial velocity, a is the acceleration of the body and s is the displacement.

Since the initial velocity is zero

$v^{2}=2as$

Making acceleration, a the subject of the formula then

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

Substituting 5 m/s for v and 2.5 m for s then

$a=\frac {5^{2}}{2\times 2.5}=5 m/s^{2}$

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3 years ago

An object takes 5 seconds to move 2 meters upward. How fast did it go?

jonny [76]

Answer:

2.5

Explanation:

5/2=2.5

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3 years ago

You push a 50 kg wooden box across a wooden floor at a constant speed of 1.0 m/s. The coefficient of kinetic friction is 0.15. N

WARRIOR [948]

Answer:

0.68 seconds

Explanation:

Data provided in the question:

Mass of the box = 50 kg

Speed of the box = 1.0 m/s

Coefficient of friction, μ = 0.15

Now,

Force applied = μmg

Here,

g is the acceleration due to gravity = 9.8 m/s²

Thus,

F = 0.15 × 50 × 9.8

= 73.5 N

Also,

Force = Mass × Acceleration

thus,

73.5 N = 50 × a

or

a = 1.47 m/s²

After doubling the speed

Final speed = 2 × Initial speed

= 2 × 1 m/s

= 2 m/s

Also,

Acceleration = [change in speed] ÷ Time

or

1.47 = [ 2 - 1 ] ÷ Time

or

Time = 1 ÷ 1.47

or

Time = 0.68 seconds

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4 years ago

The box is pushed to the right with a force of 40 Newtons and it just begins to move what is the maximum static frictional force

Vesnalui [34]

Answer:

The maximum static frictional force is 40N.

Explanation:

When an object of mass M is on a surface with a coefficient of static friction μ, there is a minimum force that you need to apply to the object in order to "break" the coefficient of static friction and be able to move the object (Called the threshold of motion, once the object is moving we have a coefficient of kinetic friction, which is smaller than the one for static friction).

This coefficient defines the maximum static friction force that we can have.

So if we apply a small force and we start to increase it, the static frictional force will be equal to our force until it reaches its maximum, and then we can move the object and now we will have frictional force.

In this case, we know that we apply a force of 40N and the object just starts to move.

Then we can assume that we are just at the point of transition between static frictional force and kinetic frictional force (the threshold of motion), thus, 40 N is the maximum of the static frictional force.

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3 years ago