All categories

1 year ago

A 10-kg dog is running with a speed of 5.0 m/s. what is the minimum work required to stop the dog in 2.40 s?

Physics

1 answer:

Tasya [4]1 year ago

3 0

<h3><u>Answer;</u></h3>

<em>Work = 125 joules </em>

<h3><u>Explanation and solution</u>;</h3>

Work is the product of force and the distance covered. Therefore, Work = force × distance.

Work is measured in joules.
Work is also a change in energy, such that work is done when energy changes, so when kinetic energy, or potential energy changes the there is work being done.

Thus; kinetic energy = work done

Kinetic energy = 1/2mv²

= 1/2 × 10× 5²

= 5 × 25

= 125 joules

Hence, work done is 125 joules.

You might be interested in

A 2.70 kg cat is sitting on a windowsill. The cat is sleeping peacefully until a dog barks at him. Startled, the cat falls from

Alchen [17]

Answer:

The speed of the cat when it hits the ground is approximately 7.586 meters per second.

Explanation:

By Principle of Energy Conservation and Work-Energy Theorem, we have that initial potential gravitational energy of the cat ( $U_{g}$ ), in joules, is equal to the sum of the final translational kinetic energy ( $K$ ), in joules, and work losses due to air resistance ( $W_{l}$ ), in joules:

$U_{g} = K +W_{l}$ (1)

By definition of potential gravitational energy, translational kinetic energy and work, we expand the equation presented above:

$m \cdot g\cdot h = \frac{1}{2}\cdot m \cdot v^{2}+W_{l}$ (2)

Where:

$m$ - Mass of the cat, in kilograms.

$g$ - Gravitational acceleration, in meters per square second.

$h$ - Initial height of the cat, in meters.

$v$ - Final speed of the cat, in meters per second.

If we know that $m = 2.70\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $h = 5.20\,m$ and $W_{l} = 120\,J$ , then the final speed of the cat is:

$v = \sqrt{\frac{2\cdot (m\cdot g\cdot h-W_{l})}{m} }$

$v = \sqrt{2\cdot g\cdot h-\frac{W_{l}}{m} }$

$v \approx 7.586\,\frac{m}{s}$

The speed of the cat when it hits the ground is approximately 7.586 meters per second.

4 0

2 years ago

A nonuniform, 80.0-g, meterstick balances when the support is placed at the 51.0-cm mark. At what location on the meterstick sho

Gnoma [55]

Answer:34 cm

Explanation:

Given

mass of meter stick m=80 gm

stick is balanced when support is placed at 51 cm mark

Let us take 5 gm tack is placed at x cm on meter stick so that balancing occurs at x=50 cm mark

balancing torque

$80\times 10^{-3}(51-50)=5\times 10^{-3}(50-x)$

$80=5(50-x)$

$80=250-5x$

$5x=170$

$x=\frac{170}{5}$

$x=34 cm$

4 0

2 years ago

The resistivity of a semiconductor can be modified by adding different amounts of impurities. A rod of semiconducting material o

zavuch27 [327]

Answer:

Explanation:

7 0

1 year ago

You travel in a circle, whose circumference is 8 kilometers, at an average speed of 8 kilometers/hour. If you stop at the same p

Schach [20]

Velocity = (displacement) / (time)

Displacement = straight-line distance between start-point and end-point

If you stop at the same point you started from, then
your displacement for the trip is zero, and your average
velocity is also zero.

5 0

1 year ago

Read 2 more answers

A projectile has an initial horizontal velocity of 15 meters per second and an initial vertical velocity of 25 meters per second

Artyom0805 [142]

Answer:

75 m

Explanation:

The horizontal motion of the projectile is a uniform motion with constant speed, since there are no forces acting along the horizontal direction (if we neglect air resistance), so the horizontal acceleration is zero.

The horizontal component of the velocity of the projectile is

$v_x = 15 m/s$

and it is constant during the motion;

the total time of flight is

t = 5 s

Therefore, we can apply the formula of the uniform motion to find the horizontal displacement of the projectile:

$d= v_x t =(15 m/s)(5 s)=75 m$

5 0

2 years ago