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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 <u>15 m</u> 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.

7 0

3 years ago

In a pool game, the cue ball, which has an initial speed of 3.0 m/s, make an elastic collision with the eight ball, which is ini

zhenek [66]

Explanation:

Given

initial speed(u)=3 m/s

mass of each ball is m

Let the cue ball is moving in x direction initially

In elastic collision Energy and momentum is conserved

Let u be the initial velocity and $v_1 , v_2$ be the final velocity of 8 ball and cue ball respectively

$\frac{mu^2}{2}+0=\frac{mv^2_1}{2}+\frac{mv^2_2}{2}$

The angle after which cue ball is deflected is given by

$\theta _1=90-40=50^{\circ}$

Conserving momentum in x direction

$mu=mv_1cos40+mv_2cos50$

$3=v_1cos40+v_2cos50$

Along Y axis

$0+0=v_1sin40-v_2sin50$

$v_1sin40=v_2sin50$

substitute the value of $v_1$

we get $v_2=1.912 m/s$

$v_1=2.27 m/s$

5 0

3 years ago

A book has a mass of 5 lb what is the mass of the book in kg

hjlf

5lbs in kgs is actually 2.268 kgs

4 0

3 years ago

The equivalent capacitance of three capacitors, each of capacitance C connected at series

katrin2010 [14]

Answer:

In Series:

1/C = 1/c1 + 1/c2 + 1/c3

Explanation:

For equivalent,

1/C = 3/c

Ce = c/3

5 0

3 years ago

A .5kg football is thrown with a velocity of 15m/s to the right. A stationary receiver catch the bail and bring it to rest in 0.

storchak [24]

<h3>Answer:</h3>

375 N

<h3>Explanation:</h3>

Topic tested: Newton's Law of motion

The question is testing on the application of Newton's second Law of motion.

We are given;

Mass of the football = 5 Kg
Initial velocity of the football, Vf = 15 m/s
Time taken to bring the ball to rest = 0.2 s
Final velocity, Vo = 0 m/s ( since the ball went to rest)

We are required to determine the force exerted to bring it to rest.

According to the Newton's second law of motion the resultant force and rate of change in momentum are directly proportion.

Therefore;

$F=\frac{(MVf-MVo)}{t}$

Thus;

$F=\frac{(5kg.5m/s)-(5kg.0m/s)}{0.2s}$

$F=375Newtons$

Force = 375 N

Hence, the force exerted on the ball by the receiver was 375 N

5 0

4 years ago