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astraxan [27]
4 years ago
14

The normal force acting on an object and the force of static friction do zero work on the object. However the reason that the wo

rk is zero is different for the two cases. Explain why each does zero work.
Physics
1 answer:
spin [16.1K]4 years ago
3 0

Answer:

<em>The normal force is perpendicular to the displacement</em>

<em>The static friction force produces no displacement</em>

Explanation:

Work Done By Special Forces

The work is a physical magnitude that measures the dot product of the force applied to an object by the displacement it produces in it.

W=\vec F\ \vec r

It can be written in its scalar version as

W=F.d.cos\theta

Being F and d the magnitudes of the force and displacement, and \theta the angle between them

If the angle is zero, the work is at maximum, it the angle is 90°, the work is zero. If the angle is between 90° and 180°, the work is negative.

The normal force acts in the vertical direction when the object is being pushed horizontally. It means the angle between the force and the displacement is 90°, thus the work is

W=N.d.cos90^o=0

The work is zero because the force and the displacement are perpendicular

The static friction force exists only when the object is being applied a force of a magnitude not large enough to produce movement, i.e. the object is at rest. If the object is moved, the friction force is still present, but it's called dynamic friction force, usually smaller than the static.

Since in this case, there is no displacement, d=0, and the work is

W=F_r(0)cos180^o=0

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nataly862011 [7]

Answer:

<em>The mass of the object is 745000 units of the sun</em>

Explanation:

We know that the centripetal force with which the stars orbit the object is represented as

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and this centripetal force is also proportional to

F_{c} = \frac{kMm}{r^{2} }

where

m is the mass of the stars

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v is the velocity of the stars = 10^6 m/s

r is the distance between the stars and the object = 10^14 m

k is the gravitational constant = 6.67 × 10^-11 m^3 kg^-1 s^-2

We can equate the two centripetal force equations to give

\frac{mv^{2} }{r} = \frac{kMm}{r^{2} }

which reduces to

v^{2} = \frac{kM}{r}

and then finally

M = \frac{rv^{2} }{k}

substituting values, we have

M = \frac{10^{14}*(10^{6})^{2}  }{6.67*10^{-11} } = 1.49 x 10^36 kg

If the mass of the sun is 2 x 10^30 kg

then, the mass of the the object in units of the mass of the sun is

==> (1.49 x 10^36)/(2 x 10^30) = <em>745000 units of sun</em>

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Answer:

<em>The velocity of the ball as it hit the ground = 19.799 m/s</em>

Explanation:

Velocity: Velocity of a body can be defined as the rate of change of displacement of the body. The S.I unit of velocity is m/s. velocity is expressed in one of newtons equation of motion, and is given below.

v² = u² + 2gs.......................... Equation 1

Where v = the final velocity of the ball, g = acceleration due to gravity, s = the height of the ball

<em>Given: s = 20 m, u = 0 m/s</em>

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<em>Therefore the velocity of the ball as it hit the ground = 19.799 m/s</em>

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