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2 years ago

When friction slows a sliding block___

Physics

1 answer:

Svetradugi [14.3K]2 years ago

6 0

Answer:

Explanation:

When friction slows a sliding block, the kinetic energy of the block is transformed into internal energy .

The frictional movement of two surfaces over one another leads to the conversion of some of their kinetic energies to another energy - heat or thermal energy. Hence, the temperatures of the objects are raised in the process.

Therefore, when a sliding block is slowed down due to friction, some of the kinetic energy of the block would be transformed into internal energy in the form of heat.

The correct option is A.

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Difference between output work and input work

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3 years ago

HELPPP PLS BESTIES A pattern of stars that has been named is called a constellation. One winter evening, Jason found the star ca

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3 years ago

A 90 kg ice skater moving at 12.0 m/s on the ice encounters a region of roughed up ice with a coefficient of kinetic friction of

balandron [24]

Answer:

The skater covers a distance of 15 m before stopping.

Explanation:

Let the distance traveled before stopping be 'd' m.

Given:

Mass of the skater (m) = 90 kg

Initial velocity of the skater (u) = 12.0 m/s

Final velocity of the skater (v) = 0 m/s (Stops finally)

Coefficient of kinetic friction (μ) = 0.490

Acceleration due to gravity (g) = 9.8 m/s²

Now, we know that, from work-energy theorem, the work done by the net force on a body is equal to the change in its kinetic energy.

Here, the net force acting on the skater is only frictional force which acts in the direction opposite to motion.

Frictional force is given as:

$f=\mu N$

Where, 'N' is the normal force acting on the skater. As there is no vertical motion, $N=mg$

∴ $f=\mu mg=0.490\times 90\times 9.8=432.18\ N$

Now, work done by friction is a negative work as friction and displacement are in opposite direction and is given as:

$W=-fd=-432.18d$

Now, change in kinetic energy is given as:

$\Delta K=\frac{1}{2}m(v^2-u^2)\\\\\Delta K=\frac{1}{2}\times 90(0-12^2)\\\\\Delta K=45\times (-144)=-6480\ J$

Therefore, from work-energy theorem,

$W=\Delta K\\\\-432.18d=6480\\\\d=\frac{6480}{432.18}\\\\d=14.99\approx 15\ m$

Hence, the skater covers a distance of 15 m before stopping.

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3 years ago

if a car that is 100 feet in front of you on route 70 west slams on the brakes while you are traveling 65 miles per hour (95 fee

fenix001 [56]

Answer:

t = 1.05 s

Explanation:

Given,

The distance between your vehicle and car, 100 ft

The constant speed of your vehicle, u = 95 ft/s

Since, the velocity is constant, a =0

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S = ut + ½ at²

Substituting the given values in the equation

100 = 95 x t

t = 100/95

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Hence, the time left for you to hit the brakes and stop before rear ending them, t = 1.05 s

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3 years ago