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How much work would it take to push a # 3 kg # weight up a # 7 m # plane that is at an incline of # pi / 6 #?

Answer 1

Work #W=102.9" "#Joules

The force #F# required to push a weight a distance #d=7" "#meters up parallel to the incline plane at #theta=pi/6=30^@# is
#F=mg*sin theta#

considering no frictional force is involve

The work #W#:
#W=F*d=mg*sin theta*d=3(9.8)*sin (pi/6)*7#
#W=102.9" "#joules

God bless....I hope the explanation is useful.

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Answer 2

The work done to push a 3 kg weight up a 7 m incline at an angle of π/6 radians can be calculated using the formula:

Work = Force × Distance × Cosine(angle)

The force required to overcome gravity along the incline is given by:

Force = Mass × Gravity × Sin(angle)

Where: Mass = 3 kg Gravity ≈ 9.8 m/s² Angle = π/6 radians

Substituting these values into the equations:

Force = 3 kg × 9.8 m/s² × Sin(π/6) ≈ 3 kg × 9.8 m/s² × 0.5

Work = Force × Distance × Cos(π/6) = (3 kg × 9.8 m/s² × 0.5) × 7 m × Cos(π/6)

Calculate the values and you'll find the work done.

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Answer 3

The work done to push the weight up the inclined plane can be calculated using the formula:

[ \text{Work} = \text{Force} \times \text{Distance} \times \cos(\theta) ]

where

  • (\text{Force}) is the force applied to push the weight up the incline,
  • (\text{Distance}) is the distance along the incline that the weight is pushed, and
  • (\theta) is the angle of the incline with respect to the horizontal.

The force required to push the weight up the incline can be calculated using the component of the weight parallel to the incline. The weight of the object is ( m \cdot g ), where ( m ) is the mass of the object and ( g ) is the acceleration due to gravity (approximately ( 9.81 , \text{m/s}^2 )). The component of the weight parallel to the incline is ( m \cdot g \cdot \sin(\theta) ).

Plugging in the values, we get:

[ \text{Force} = 3 , \text{kg} \times 9.81 , \text{m/s}^2 \times \sin\left(\frac{\pi}{6}\right) ]

[ \text{Force} = 3 \times 9.81 \times \frac{1}{2} ]

[ \text{Force} = 14.715 , \text{N} ]

Now, we can calculate the work done:

[ \text{Work} = 14.715 , \text{N} \times 7 , \text{m} \times \cos\left(\frac{\pi}{6}\right) ]

[ \text{Work} = 14.715 \times 7 \times \frac{\sqrt{3}}{2} ]

[ \text{Work} = 14.715 \times 7 \times 0.866 ]

[ \text{Work} = 89.48 , \text{J} ]

So, it would take approximately 89.48 joules of work to push a 3 kg weight up a 7 m plane that is at an incline of ( \frac{\pi}{6} ) radians.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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