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miv72 [106K]
3 years ago
14

You want to push a 71-kg box up a 16° ramp. The coefficient of kinetic friction between the ramp and the box is 0.23. With what

magnitude force parallel to the ramp should you push on the box so that it moves up the ramp at a constant speed?
Physics
1 answer:
ycow [4]3 years ago
8 0

Answer:

<em>345.60N</em>

Explanation:

The magnitude of the force parallel to the ramp that will push the box up the ramp are the frictional force(Ff) and the moving force Fm

Total force = Ff + Fm

Ff = ηR

Ff = ηmgcosθ

Fm = Wsinθ

Fm = mgsinθ

Total force = ηmgcosθ + mgsinθ

Total force = mg(ηcosθ + sinθ)

Given

mass m = 71kg

acceleration due to gravity g = 9.8m/s²

angle of inclination θ = 16°

coefficient of friction η = 0.23

Substitute into the formula;

Total force = 71(9.8)(0.23cos16 + sin16)

Total force = 695.8(0.2211+0.2756)

Total force = 695.8(0.4967)

Total force = 345.60N

<em>Hence the magnitude force parallel to the ramp that you should push on the box so that it moves up the ramp at a constant speed is 345.60N</em>

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Answer:

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Explanation:

Given:

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0.153 × 0.7 + 0.308 × -2.16 = 0.153 × V1 + 0.308 × V2

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0.153V1 + 0.308V2 = -0.55818. i

For the velocities,

v1 - v2 = -(V1 - V2)

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Maximum height reached by the ball is 32 meters.

Explanation:

It is given that,

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t is the time taken

s is the height attained as a function of time.

Maximum height achieved can be calculated as :

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Answer:

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v_{2}=\frac{(65.94-66)ft}{(2.1-2)s}=-0.6ft/s

v_{3}=\frac{(66.0084-66)ft}{(2.01-2)s}=0.84ft/s

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Explanation:

From the exercise we got the ball's equation of position:

y=65t-16t^{2}

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v=\frac{y_{2}-y_{1}  }{t_{2}-t_{1}  }

Being said that, we need to find the ball's position at t=2, t=2.5, t=2.1, t=2.01, t=2.001

y_{t=2}=65(2)-16(2)^{2} =66ft

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--

y_{t=2.1}=65(2.1)-16(2.1)^{2} =65.94ft

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Physical property is defined as the property of a substance which becomes evident during physical change in which there is alteration in shape, size etc. No new substance gets formed during physical change.

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