A sailboat sits tied up at the dock.

In the figure below the pulley is a solid disk of mass M and radius R with rotational inertia MR 2/2. Two blocks one of mass m a

matrenka [14]

Assuming you are looking for the acceleration a:

1. $m_1a = T_1 -m_1g$
2. $m_2a = m_2g - T_2$
where T is the tension and a is the acceleration of the blocks. The acceleration of the two blocks and the acceleration of the pulley must be equal.

The torque on the pulley is given by:
3. $\tau = \overrightarrow r \times \overrightarrow F = (T_2 - T_1)R = I\alpha = \frac{1}{2} MR^2 \frac{a}{R}$
where $I = \frac{1}{2} mR^2$ and $a = \alpha R$ .

Combining the three equations:
$T_2 - T_1 = \frac{1}{2} Ma \\ m_2g - m_2a -m_1g - m_1a = (m_2-m_1)g - (m_1 + m_2)a = \frac{1}2}Ma \\ \\ a = \frac{(m_2 - m_1)g}{m_1 + m_2 + \frac{1}{2}M }$

6 0

3 years ago

Soil can best be described as the

Rom4ik [11]

<span>The correct option is D. Soil can best be described as the loose covering of weathered rocks and decaying organic matter. There are different types of soil, the type of soil formed depends majorly on the type of parent rock from which the soil is formed and the amount of organic matter present in the soil.</span>

5 0

3 years ago

Using energy considerations, calculate the average force (in N) a 62.0 kg sprinter exerts backward on the track to accelerate fr

slava [35]

Answer:

69.68 N

Explanation:

Work done is equal to change in kinetic energy

W = ΔK = Kf - Ki = $\frac{1}{2} mv^{2} _{f} - \frac{1}{2} mv^{2} _{i}$

W = $F_{total} .d$

where m = mass of the sprinter

vf = final velocity

vi = initial velocity

W = workdone

kf = final kinetic energy

ki = initial kinetic energy

d = distance traveled

Ftotal = total force

vf = 8m/s

vi= 2m/s

d = 25m

m = 60kg

inserting parameters to get:

W = ΔK = Kf - Ki = $\frac{1}{2} mv^{2} _{f} - \frac{1}{2} mv^{2} _{i}$

$F_{total} .d =\frac{1}{2} mv^{2} _{f} - \frac{1}{2} mv^{2} _{i}$

$F_{total} = \frac{\frac{1}{2} mv^{2} _{f} - \frac{1}{2} mv^{2} _{i}}{d}$

$F_{total=} \frac{\frac{1}{2} X 62 X6^{2} -\frac{1}{2} X 62 X2^{2} }{25}$

= 39.7

we know that the force the sprinter exerted F sprinter, the force of the headwind Fwind = 30N

$F_{sprinter} = F_{total} + F_{wind} = 39.7 + 30 = 69.68 N$

7 0

3 years ago

Read 2 more answers

Your roommate drops your wallet down to you from the third-floor window of your apartment, which is 11.5 m from the ground. What

Aneli [31]

Answer:

15 m/s

Explanation:

Using the law of conservation of energy, potential energy equals kinetic energy hence

$mgh=0.5mv^{2}$

Therefore

$v=\sqrt{2gh}$

where g is the acceleration due to gravity, m is the mass of the object, h is the height and v is the speed of the wallet

Taking g as 9.81 then

$v=\sqrt{2\times 9.8\times 11.5}=15.02098532 m/s\approx 15 m/s$

6 0

3 years ago

A solid block, with a mass of 0.15kg, on a frictionless surface is pushed directly onto a horizontal spring, with a spring const

iren [92.7K]

Answer:

16.1 m/s

Explanation:

We can solve the problem by using the law of conservation of energy.

At the beginning, the spring is compressed by x = 35 cm = 0.35 m, and it stores an elastic potential energy given by

$U=\frac{1}{2}kx^2$

where k = 316 N/m is the spring constant. Once the block is released, the spring returns to its natural length and all its elastic potential energy is converted into kinetic energy of the block (which starts moving). This kinetic energy is equal to

$K=\frac{1}{2}mv^2$

where m = 0.15 kg is the mass of the block and v is its speed.

Since the energy must be conserved, we can equate the initial elastic energy of the spring to the final kinetic energy of the block, and from the equation we obtain we can find the speed of the block:

$\frac{1}{2}kx^2=\frac{1}{2}mv^2\\v=\sqrt{\frac{kx^2}{m}}=\sqrt{\frac{(316 N/m)(0.35 m)^2}{0.15 kg}}=16.1 m/s$

4 0

3 years ago