How much work is done when a 48-kg stack of books are lifted 2m with a net force of 25 N ?

The Atwood’s machine shown consists of two blocks of mass m1 and m2 that are connected by a light string that passes over a pull

Talja [164]

(A) For the system consisting of the two blocks, the change in the kinetic energy of the system is equal to work done by gravity on the system.

(D) For the system consisting of the two blocks, the pulley and the Earth, the change in the total mechanical energy of the system is zero.

<h3 /><h3>The given parameters:</h3>

Mass of block 1 = m1
Mass of block 2, = m2
Height of block 1 above the ground, = h1
Height of block 2 above the ground = h2

The total initial mechanical energy of the two block system is calculated as follows;

$m_1gh_1 + \frac{1}{2} m_1v_1_i^2 = m_2gh_2 + \frac{1}{2} m_2v_2_i^2\\\\m_1gh_1 + 0 = m_2gh_2 + 0\\\\m_1gh_1 = m_2gh_2\\\\m_1gh_1 - m_2gh_2 = 0$

When the block m2 reaches the ground the block m1 attains maximum height and the total mechanical energy at this point is given as;

$m_1g(h_1 + h_2) + K.E_1 = \frac{1}{2}m_2v_{max}^2 + P.E_2\\\\m_1g(h_1 + h_2 ) -PE_2 = \frac{1}{2}m_2v_{max}^2 - K.E_1\\\\m_1g(h_1 + h_2 ) - 0= \frac{1}{2}m_2v_{max}^2 - 0\\\\m_1g(h_1 + h_2 ) = \frac{1}{2}m_2v_{max}^2\\\\W = \frac{1}{2}m_2v_{max}^2$

Thus, we can conclude the following before the block m2 reaches the ground;

For the system consisting of the two blocks, the pulley and the Earth, the change in the total mechanical energy of the system is zero.
For the system consisting of the two blocks, the change in the kinetic energy of the system is equal to work done by gravity on the system.

Learn more about conservation of mechanical energy here: brainly.com/question/332163

5 0

3 years ago

A mine car (mass=390 kg) rolls at a speed of 0.50 m/s on a horizontal track, as the drawing shows. A 250-kg chunk of coal has a

Nady [450]

Answer:

v=0.60 m/s

Explanation:

Given that

m ₁= 390 kg ,u ₁= 0.5 m/s

m₂ = 250 kg ,u₂ = 0.76 m/s

As we know that if there is no any external force on the system the total linear momentum of the system will be conserve.

Pi = Pf

m ₁u ₁+m₂u₂ = (m₂ + m ₁ ) v

Now putting the values in the above equation

390 x 0.5 + 250 x 0.76 = (390 + 250 ) v

$v=\dfrac{390\times 0.5+250\times 0.76}{390+250}\ m/s$

v=0.60 m/s

Therefore the velocity of the system will be 0.6 m/s.

8 0

3 years ago

Q) A farmer moves along the boundary of a

Alja [10]

The answer will be 8 because kedks

4 0

3 years ago

A mass-spring system has k = 56.8 N/m and m = 0.46 kg.

andriy [413]

Answer:

A. $\omega=11.1121\ rad.s^{-1}$

B. $f=1.7685\ Hz$

C. $T=0.5654\ s$

Explanation:

Given:

spring constant, $k=56.8\ N.m^{-1}$

mass attached, $m=0.46\ kg$

A)

for a spring-mass system the frequency is given as:

$\omega=\sqrt{\frac{k}{m} }$

$\omega=\sqrt{\frac{56.8}{0.46}}$

$\omega=11.1121\ rad.s^{-1}$

B)

frequency is given as:

$f=\frac{\omega}{2\pi}$

$f=\frac{11.1121}{2\pi}$

$f=1.7685\ Hz$

C)

Time period of a simple harmonic motion is given as:

$T=\frac{1}{f}$

$T=0.5654\ s$

7 0

3 years ago

A loop of wire is in a magnetic field such that its axis is parallel with the field direction. Which of the following would resu

ahrayia [7]

Answer:

All the given options will result in an induced emf in the loop.

Explanation:

The induced emf in a conductor is directly proportional to the rate of change of flux.

$emf = -\frac{d \phi}{dt} \\\\where;\\\\\phi \ is \ magnetic \ flux\\\\\phi = BA\ cos \theta$

where;

A is the area of the loop

B is the strength of the magnetic field

θ is the angle between the loop and the magnetic field

<em>Considering option </em><em>A</em>, moving the loop outside the magnetic field will change the strength of the magnetic field and consequently result in an induced emf.

<em>Considering option </em><em>B</em>, a change in diameter of the loop, will cause a change in the magnetic flux and in turn result in an induced emf.

Option C has a similar effect with option A, thus both will result in an induced emf.

Finally, <em>considering option</em> D, spinning the loop such that its axis does not consistently line up with the magnetic field direction will<em> </em>change the angle<em> </em>between the loop and the magnetic field. This effect will also result in an induced emf.

Therefore, all the given options will result in an induced emf in the loop.

4 0

3 years ago