15.0 kg mass is displaced 3.00 m south and then 4.00 m west by a 10.0 N force. What is the total work done on the object?

Alchen [17]

courts have many properties including Mass electric charge and color their Mass tells you something about how much matter they are made of or how heavy they would be if you could wear them up charm and top quarks have a positive 2/3 charge

7 0

4 years ago

A hoop and a solid disc are relased from rest

Murljashka [212]

Answer:

1) The hoop and a solid disc rolling without slipping down an incline plane.

Their final velocities are proportional to their moment of inertia.

The condition for moment of inertia: v = ωR

We will use conservation of energy.

For the hoop:

$K_1 + U_1 = K_2 + U_2\\0 + m_hgh = \frac{1}{2}m_hv_h^2 + \frac{1}{2}I\omega_h^2 + 0$

They are released from rest, so their initial kinetic energy is zero. And when they reach the bottom, their final potential energy is also zero.

The moment of inertia of a hoop is

$I_h = m_hR^2$

Let's continue with the energy equations:

$m_h gh = \frac{1}{2}m_hv_h^2 + \frac{1}{2}(m_hR^2)(\frac{v_h^2}{R^2})\\m_hgh = \frac{1}{2}m_hv_h^2 + \frac{1}{2}m_hv_h^2\\m_hgh = m_hv_h^2\\v_h = \sqrt{gh}$

Similarly for the solid disk with a moment of inertia of (1/2)mR^2:

$K_1 + U_1 = K_2 + U_2\\m_dgh = \frac{1}{2}m_dv_d^2 + \frac{1}{2}I_d\omega_d^2\\m_dgh = \frac{1}{2}m_dv_d^2 + \frac{1}{2}(\frac{1}{2}m_dR^2)(\frac{v_d^2}{R^2})\\m_dgh = \frac{1}{2}m_dv_d^2 + \frac{1}{4}m_dv_d^2\\m_dgh = \frac{3}{4}m_dv_d^2\\v_d = \sqrt{\frac{4gh}{3}}$

Comparing the final velocities, we can conclude that the solid disk reaches the bottom first.

2) The angular acceleration of the pebble is equal to the angular acceleration of the tire, since they stuck together. We can deduce the angular acceleration of the tire from the linear acceleration of the bicycle.

The kinematics equations states that

$v = v_0 + at\\4.47 = 0 + 2a\\a = 2.235 ~m/s^2$

where a is the linear acceleration.

The relation with the angular and linear acceleration is

$a = \alpha R$

where R is the radius of the tire. Since it is not given in the question, we will leave it as R.

The angular acceleration of the small pebble is

$\alpha = 2.235/R ~m/s^2$

4 0

4 years ago

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9. Luke is an astronaut that moves his position in the space shuttle by climbing a

BartSMP [9]

Explanation:

Remember that power is defined as how much Work is done per unit of time.

$P = \frac{dW}{dt}$

Work is defined as the amount of force applied across a certain distance.

$W = F*d$

Since in both cases of climbing the ladder (on Earth and the moon), Luke coveres the same amount of distance in the same amount of time, we are only left with one difference between the two cases - gravity.

If you were to carry your backpack on the moon with the same load of text books, it would take less force to pick it up on the moon. Therefore, Luke expends less effort on the environment with less gravity - the moon.

To find the difference factor - you would want to divid the gravitational contants between earth and the moon.

6 0

3 years ago

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A solid sphere is rolling on a surface. what fraction of its total kinetic energy is in the form of rotational kinetic energy ab

drek231 [11]

Total KE = KE (rotational) + KE (translational) Moment of inertia of sphere is I = (2/5)mr^2 So KE (rotational) = (1/2) x I x w^2 = (1/2) x (2/5)mr^2 x w^2 = (1/5) x m x r^2 x w^2 KE (translational) = (1/2) x m x v^2 = (1/2) x m x (rw)^2 = (1/2) x m x r^2 x w^2 Hence KE = (1/5) x m x r^2 x w^2 + (1/2) x m x r^2 x w^2 = m x r^2 x w^2 ((1/5) + (1/2)) KE = (7/10) m x r^2 x w^2 Calculating the fraction of rotational kinetic energy to total kinetic energy, = rotational kinetic energy / total kinetic energy = (1/5) x m x r^2 x w^2 / (7/10) m x r^2 x w^2 = (1/5) / (7/10) = 2 / 7 The answer is 2 / 7

7 0

3 years ago

You are an engineer helping to design a roller coaster that carries passengers down a steep track and around a vertical loop. Th

vova2212 [387]

Answer:

h >5/2r

Explanation:

This problem involves the application of the concepts of force and the work-energy theorem.

The roller coaster undergoes circular motion when going round the loop. For the rider to stay in contact with the cart at all times, the roller coaster must be moving with a minimum velocity v such that at the top the rider is in a uniform circular motion and does not fall out of the cart. The rider moves around the circle with an acceleration a = v²/r. Where r = radius of the circle.

Vertically two forces are acting on the rider, the weight and normal force of the cart on the rider. The normal force and weight are acting downwards at the top. For the rider not to fall out of the cart at the top, the normal force on the rider must be zero. This brings in a design requirement for the roller coaster to move at a minimum speed such that the cart exerts no force on the rider. This speed occurs when the normal force acting on the rider is zero (only the weight of the rider is acting on the rider)

So from newton's second law of motion,

W – N = mv²/r

N = normal force = 0

W = mg

mg = ma = mv²/r

mg = mv²/r

v²= rg

v = √(rg)

The roller coaster starts from height h. Its potential energy changes as it travels on its course. The potential energy decreases from a value mgh at the height h to mg×2r at the top of the loop. No other force is acting on the roller coaster except the force of gravity which is a conservative force so, energy is conserved. Because energy is conserved the total change in the potential energy of the rider must be at least equal to or greater than the kinetic energy of the rider at the top of the loop

So

ΔPE = ΔKE = 1/2mv²

The height at the roller coaster starts is usually higher than the top of the loop by design. So

ΔPE =mgh - mg×2r = mg(h – 2r)

2r is the vertical distance from the base of the loop to the top of the loop, basically the diameter of the loop.

In order for the roller coaster to move smoothly and not come to a halt at the top of the loop, the ΔPE must be greater than the ΔKE at the top.

So ΔPE > ΔKE at the top. The extra energy moves the rider the loop from the top.

ΔPE > ΔKE

mg(h–2r) > 1/2mv²

g(h–2r) > 1/2(√(rg))²

g(h–2r) > 1/2×rg

h–2r > 1/2×r

h > 2r + 1/2r

h > 5/2r

5 0

3 years ago

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