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HACTEHA [7]
3 years ago
7

The pulley system below uses a gasoline engine to raise a drill head up through a smooth drill pipe. The engine provides a const

ant force of 900 N. The drill head has a mass of 80 kg. Find the power that is being consumed by the motor (provided by the gasoline) at the instant the drill head has been raised 5 m., starting from rest. The engine has an efficiency of 0.65.
Physics
1 answer:
polet [3.4K]3 years ago
6 0

NB: The diagram of the pulley system is not shown but the information provided is sufficient to answer the question

Answer:

Power = 2702.56 W

Explanation:

Let the power consumed be P

Energy expended = E = mgh

height, h = 5 m

E = 80 * 9.8 * 5

E = 3920 J

Power = \frac{Energy}{time}

To calculate the time, t

From F = ma

F = 900 N

900 = 80 a

a = 900/80

a = 11.25 m/s²

From the equation of motion, s = ut + 0.5at^{2}

The drill head starts from rest, u = 0 m/s

5 = 0 * t + (0.5*11.25*t^{2} )\\5 = 5.625t^{2}\\t^{2} = 5/5.625\\t^{2} = 0.889\\t = 0.943 s

Power, P = E/t

P = 3920/0.0.943

P = 4157.79 W

But Efficiency, E = 0.65

P = 0.65 * 4157.79

Power = 2702.56 W

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Two identical copper blocks are connected by a weightless, unstretchable cord through a frictionless pulley at the top of a thin
likoan [24]

Answer:

13.6 N

Explanation:

Since one side of the wedge is vertical and the wedge makes and angle of 33 with the horizontal, the angle between the weight of the copper block on the incline and the incline is thus 90 - 33 = 57.

Let M be the mass of the block that hangs, m be the mass of the block on the incline and T be the tension in the weightless unstretchable cord.

We assume the motion is downwards in the direction of the hanging block, M.

We now write equations of motion for each block.

So

Mg - T = Ma    (1) and T - mgcos57 - F = ma where F is the frictional force on the block on the incline and a is their acceleration.

Now, since both blocks do not move, a = 0.

So, Mg - T = M(0) = 0     and T - mgcos57 - F = m(0) = 0

Mg - T = 0    (3) and T - mgcos57 - F = 0 (4)

From (3), T = Mg

Substituting T into (4), we have

T - mgcos57 - F = 0

Mg - mgcos57 - F = 0

So, Mg - mgcos57 = F  

F = Mg - mgcos57

F = (M - mcos57)g

Since g = acceleration due to gravity = 9.8 m/s², and M = 2.94 kg and m = 2.85 kg.

We find F, thus

F = (2.94 kg - 2.85 kgcos57)9.8 m/s²

F = (2.94 kg - 2.85 kg × 0.5446)9.8 m/s²

F = (2.94 kg - 1.552 kg)9.8 m/s²

F = (1.388 kg)9.8 m/s²

F = 13.6024 kgm/s²

F ≅ 13.6 N

6 0
3 years ago
if a torque of 55.0 N/m is required and the largest force that can be exerted by you is 135 N what is th e length of the lever a
Whitepunk [10]

Answer:

r=0.41m

Explanation:

Torque is defined as the cross product between the position vector ( the lever arm vector connecting the origin to the point of force application) and the force vector.

\tau=r\times F

Due to the definition of cross product, the magnitude of the torque is given by:

\tau=rFsin\theta

Where \theta is the angle between the force and lever arm vectors. So, the length of the lever arm (r) is minimun when sin\theta is equal to one, solving for r:

r=\frac{\tau}{F}\\r=\frac{55\frac{N}{m}}{135N}\\r=0.41m

7 0
2 years ago
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