All categories

2 years ago

Determine the changeKf−Ki in the kinetic energy of the block as it moves from xi = -4.00 cm to xf = 6.00 cm .

Physics

1 answer:

zepelin [54]2 years ago

5 0

Answer:

A) W1 = 10.0 J

B) W2 = -2.00 J

C) Wnet = 8.0 J

D) kf - ki = 8.0 J

Explanation:

Hi! I´ve found the complete question on the web:

Two forces, of magnitudes F1 = 100 N and F2 = 20.0 N, act in opposite directions on a block, which sits atop a friction-less surface. Initially, the center of the block is at position xi = -4.00 cm. At some later time, the block has moved to the right, and its center is at a new position, xf = 6.00 cm.

A) Find the work W1 done on the block by the force of magnitude F1 = 100 N as the block moves from xi = -4.00 cm to xf = 6.00 cm.

B) Find the work W2 done by the force of magnitude F2 = 20.0 N as the block moves from xi = -4.00 cm to xf = 6.00 cm.

C) What is the net work Wnet done on the block by the two forces?

D) Determine the change Kf-Ki in the kinetic energy of the block as it moves from xi = -4.00 cm to xf = 6.00 cm.

A) The equation of work is the following:

W = F · d

Where:

W = work.

F = applied force.

d = traveled distance.

The traveled distance from -4.00 cm to 6.00 is 10.00 cm (final position - initial position = 6.00 cm - (-4.00 cm) = 10.00 cm = 0.1000 m).

Then, the work done by F1 is the following:

W1 = F1 · d

W1 = 100 N · 0.1000 m = 10.0 J

B) Similarly, we can calculate the work done by the force F2:

W2 = F2 · d

Notice that since the force is applied in the direction opposite to the displacement, the work is negative:

W2 = - 20.0 N · 0.100 m = -2.00 J

C) The net work can be calculated by adding the work done by all the forces on the object. In this case:

Wnet = W1 + W2

Wnet = 10.0 J - 2.00 J = 8.0 J

D) According to the work-energy theorem, the net work on an object is equal to its change in kinetic energy:

Wnet = ΔKE = kf - ki = 8.0 J

This means that the total work done on the block increased its kinetic energy by 8.0 J.

You might be interested in

Fiber optic (FO) cables are based upon the concept of total internal reflection (TIR), which is achieved when the FO core and cl

kozerog [31]

Answer:

False

Explanation:

Though fiber active cable is based on the concept of internal reflection but it is achieved by refractive index which transmit data through fast traveling pulses of light. It has a layer of glass and insulating casing called “cladding,”and this is is wrapped around the central fiber thereby causing light to continuously bounce back from the walls of the Cable.

7 0

2 years ago

The U.S. Department of Energy had plans for a 1500-kg automobile to be powered completely by the rotational kinetic energy of a

navik [9.2K]

Answer:

230

Explanation:

$\omega$ = Rotational speed = 3600 rad/s

I = Moment of inertia = 6 kgm²

m = Mass of flywheel = 1500 kg

v = Velocity = 15 m/s

The kinetic energy of flywheel is given by

$K=\dfrac{1}{2}I\omega^2\\\Rightarrow K=\dfrac{1}{2}6\times 3600^2\\\Rightarrow K=38880000\ J$

Energy used in one acceleration

$K=\dfrac{1}{2}mv^2\\\Rightarrow K=\dfrac{1}{2}1500\times 15^2\\\Rightarrow K=168750\ J$

Number of accelerations would be given by

$n=\dfrac{38880000}{168750}\\\Rightarrow n=230.4$

So the number of complete accelerations is 230

8 0

2 years ago

An archer draws her bow and stores 34.8 J of elastic potential energy in the bow. She releases the 63 g arrow, giving it an init

elena-14-01-66 [18.8K]

Answer:

Approximately $71\%$ .

Explanation:

The formula for the kinetic energy $\rm KE$ of an object is:

$\displaystyle \mathrm{KE} = \frac{1}{2}\, m \cdot v^2$ ,

where

$m$ is the mass of that object, and
$v$ is the speed of that object.

Important: Joule ( $\rm J$ ) is the standard unit for energy. The formula for $\rm KE$ requires two inputs: mass and speed. The standard unit of mass is $\rm kg$ while the standard unit for speed is $\rm m \cdot s^{-1}$ . If both inputs are in standard units, then the output (kinetic energy) will also be in the standard unit (that is: joules,

Convert the unit of the arrow's mass to standard unit:

$m = 63\; \rm g = 0.063\; \rm kg$ .

Initial $\rm KE$ of this arrow:

$\begin{aligned}\mathrm{KE} &= \frac{1}{2} \, m \cdot v^2 \\ &= \frac{1}{2}\times 0.063\; \rm kg \times \left(\rm 28 \; m \cdot s^{-1}\right)^2 \\ &\approx 24.696\; \rm J\end{aligned}$ .