Please solve this physics problem........?

Question

An inclined plane 10m long has one end raised 10m.If a force of 500N is needed to push a box of weight of 800N ,Find the following...
a. efficiency
b. work input
c. work output
d. Mechanical advantage

Raised by 4m

Answer 1

the goal is to make the box reach incline top by the most economical means. The better mode is to lift the box vertically against (mg force) and do work as (mgh). this route will offer only air resistance, but not friction (kinematic) because no contact surfaces will be involved.
so
work output = m g h = work done vertically (possibility)
work output = 800*4 = 3200 Joule ---(1)
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by going along the plane, the gravitational force that will oppose the motion is rather (mg sin p) where p is angle of incline >> sin p = 4/10 >> p = 23.578 deg

work done along plane = mg sin (23.578)*L + f*L (against friction)
work done (plane) = 800*(4/10)*L + f*L
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equation of motion on incline
net force = 500 = mg *(4/10) + f
f = friction force = 500 - 800*4/10 = 180 N
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work done (plane) = 800*(4/10)*L + 180*L = 3200+1800
so input work (plane) = 5000 Joule ---- (2)
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efficiency of plane = output /input = 3200/5000 = 0.64
or 64% efficiency
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mech advantag = MA = 1/sin p = 2.5 (using sin p= 4/10)

Answer 2

The velocity ratio of the plane is = [height of the plane / length of the plane]
= 10 / 4 = 2.5

The mechanical advantage of the plane is [Load / effort] = 800 / 500 = 1.6

Efficiency of the plane, η = [Mechanical advantage / velocity ratio]

= 1.6 / 2.5 = 0.64 = 64%.

That is for 100 Joule of input, the output will be 64 J.

Input work = applied force x distance moved = 500 x 10 = 5000J.

Out put work = 800 x 4 = 3200 J.

η can also be calculated using Work out put / work input = 3200 / 5000 = 64%.
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Note the following.

If there were no friction, for ideal machine, the mechanical advantage will be equal to velocity ratio.

Velocity ratio is constant for all inclined planes when power is applied parallel to the plane.

But Mechanical advantage depends upon the friction involved in the machine and hence it will differ for different planes even though the power is applied parallel to the plane.


The advantage of using the GIVEN inclined plane is:

If we apply 1 N the plane will make available a force of 1.6 N.

If we apply 500N, it provides 1.6 times 500 = 800 N.

If there were no friction,
then the m.a will be 2.5 and efficiency will be 100%.

If we apply 1N it will provide 2.5 N

if we apply 320 N the plane will provide 2.5 x 320 = 800 N

As there is no friction, the mechanical advantage equals the velocity ratio and


Efficiency will be = 320 x10 / 800x 4 = 100 %.
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Answer 3

The component of weight acting parallel to the plane is 4*800/10 = 320
a. efficiency = 320/500 = 0.64, or 64%
b. work input = 500*10 = 5,000 N-m
c. work output = 800*4 = 3,200 N-m
d. mechanical advantage = 10/4 = 2.5