Physics, 20.07.2019 08:40 faith365

1000 kg elevator accelerates upward at 1.0 m/s 2 for 10 m, starting from rest. a. how much work does gravity do on the elevator? b. how much work does the tension in the elevator cable do on the elevator? c. use the work-kinetic energy theorem to find the kinetic energy of the elevator as it reaches 10 m. d. what is the speed of the elevator as it reaches 10 m?

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Answer from: jerali

a. The amount of work gravity does on the elevator is 98,000 Joules.

b. The amount of work the tension in the elevator cable does on the elevator is 108,000 Joules.

c. The kinetic energy of the elevator as it reaches 10 meter is 10,000 Joules.

d. The speed of the elevator as it reaches 10 meter is 4.47 m/s.

Given the following data:

Mass of elevator = 1000 kgAcceleration = 1.0 m/s^2

Displacement = 10 meters

We know that the acceleration due to gravity (g) of an object on planet Earth is equal to 9.8 m/s^2

a. To find how much work gravity does on the elevator:

Mathematically, the work done by an object is given by the formula;

$Work\;done = Force \times displacement$

First of all, we would determine the force acting on the elevator due to gravity.

$Force = mg\\\\Force = 1000 \times 9.8$

Force = 9800 Newton

$Work\;done = 9800 \times 10$

Work done = 98000 Joules.

b. To find how much work the tension in the elevator cable does on the elevator;

$Tension = m(a+g)\\\\Tension = 1000(1+9.8)\\\\Tension = 1000(10.8)$

Tension = 10,800 Newton.

$Work\;done = Tensional \;force \times displacement$

$Work\;done = 10800 \times 10$

Work done = 108,000 Joules.

c. To find the kinetic energy of the elevator as it reaches 10 m, by using the work-kinetic energy theorem:

$Work\;done = K.E +P.E$

$P.E = mgh\\\\P.E = 1000\times9.8\times10$

Potential energy = 98000 Joules.

$108000=K.E + 98000\\\\K.E = 108000-98000$

K.E = 10,000 Joules.

d. To find the speed of the elevator as it reaches 10 m:

$K.E = \frac{1}{2} mv^2\\\\2K.E =mv^2\\\\v = \sqrt{\frac{2K.E}{m} } \\\\v = \sqrt{\frac{2\times 10000}{1000}} \\\\v = \sqrt{20}$

Speed, v = 4.47 m/s

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Answer from: jaygamer37

a) -98,000 J

The force of gravity is equal to the weight of the elevator:

F_g = mg=(1000 kg)(9.8 m/s^2)=9,800 N

The work done by gravity on the elevator is equal to the product between the weight of the elevator, the displacement and the angle between the direction of the force of gravity and the displacement:

$W=F_g d cos \theta = (9,800 N)(10 m)(cos 180^{\circ})=-98,000 J$

And the work done is negative because the force of gravity (downward) is in the opposite direction to the displacement (upward)

b) +108,000 J

We need to find the tension in the cable first.

Newton's second law states that the resultant of the forces acting on the elevator is equal to the product between the mass of the elevator and its acceleration:

T-F_g = ma

where T is the tension in the cable and a = 1.0 m/s^2 is the acceleration. Substituting and re-arranging the equation, we find

T=ma+F_g = (1000 kg)(1.0 m/s^2)+9,800 N=10,800 N

So, the work done by the tension in the cable is

$W=Td cos \theta = (10,800 N)(10 m)(cos 0^{\circ})=108,000 J$

And this time the work done is positive, because the tension in the cable is in the same direction of the displacement of the elevator.

c) 10,000 J

The net work done on the elevator is:

$W_{net}=108,000 J-98,000 J=10,000 J$

And this is equal to the gain in kinetic energy of the elevator:

$W_{net}=K_f - K_i$

But the initial kinetic energy is zero, because the elevator starts from rest, so Ki=0 and the final kinetic energy is

K_f = 10,000 J

d) 4.47 m/s

The final kinetic energy of the elevator is:

$K_f = \frac{1}{2}mv^2 = 10,000 J$

where m is the elevator's mass and v is the final speed of the elevator.

By re-arranging the equation, we find

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2\cdot 10,000 J}{1000 kg}}=4.47 m/s$

Answer from: Quest

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Answer from: Quest

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1000 kg elevator accelerates upward at 1.0 m/s 2 for 10 m, starting from rest. a. how much work does...

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