Ask question

Ask question

All categories

2 years ago

7

A mass of 0.4 kg hangs motionless from a vertical spring whose length is 0.76 m and whose unstretched length is 0.41 m. Next the

mass is pulled down to where the spring has a length of 1.01 m and given an initial speed upwards of 1.6 m/s. What is the maximum length of the spring during the motion that follows?

1 answer:

Blizzard [7]2 years ago

6 0

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

= 1.256 m

<h3><u>Explanation;</u></h3>

We can start by finding the spring constant

F = k*y

Therefore; k = F/y = m*g/y

= 0.40kg*9.8m/s^2/(0.76 - 0.41)

= 11.2 N/m

Energy is conserved

Let A be the maximum displacement

Therefore; 1/2*k*A^2 = 1/2*k*(1.20 - 0.41)^2 + 1/2*m*v^2

Thus; A = sqrt((1.20 - 0.55)^2 + m/k*v^2)

= sqrt((1.20 -0.55)^2 + 0.40/9.8*1.6^2)

= 0.846 m

Thus; the length will be 0.41 + 0.846 = 1.256 m

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 car of mass 1100kg moves at 24 m/s. What is the braking force needed to bring the car to a halt in 2.0 seconds? N

LenaWriter [7]

13200N

Explanation:

Given parameters:

Mass = 1100kg

Velocity = 24m/s

time = 2s

unknown:

Braking force = ?

Solution:

The braking force is the force needed to stop the car from moving.

Force = ma = $\frac{mv}{t}$

m is the mass of the car

v is the velocity

t is the time taken

Force = $\frac{1100 x 24}{2}$ = 13200N

Learn more:

Force brainly.com/question/4033012

#learnwithBrainly

8 0

2 years ago

The sound level at 1.0 m from a certain talking person talking is 60 dB. You are surrounded by five such people, all 1.0 m from

Hunter-Best [27]

Answer:

66.98 db

Explanation:

We know that

$L_T=L_S+10log(n)$

L_T= Total signal level in db

n= number of sources

L_S= signal level from signal source.

$L_T=60+10 log(5)$

= 66.98 db

7 0

1 year ago

In an attempt to impress its friends, an acrobatic beetle runs and jumps off the bottom step of a flight of stairs. The step is

Aleks04 [339]

Answer:

0.3677181864 m

Explanation:

u = Velocity = 1.5 m/s

$\theta$ = Angle = 20°

y = -20 cm

Velocity components

$u_x=ucos\theta\\\Rightarrow u_x=1.5cos20\\\Rightarrow u_x=1.40953\ m/s$

$u_y=usin\theta\\\Rightarrow u_y=1.5sin20\\\Rightarrow u_y=0.51303\ m/s$

Acceleration components

$a_x=0$

$a_y=-9.81\ m/s^2$

$y=u_yt+\dfrac{1}{2}a_yt^2\\\Rightarrow -0.2=0.51303\times t+\dfrac{1}{2}\times -9.81t^2\\\Rightarrow 4.905t^2-0.51303t-0.2=0$

$t=\frac{-\left(-0.51303\right)+\sqrt{\left(-0.51303\right)^2-4\cdot \:4.905\left(-0.2\right)}}{2\cdot \:4.905}, \frac{-\left(-0.51303\right)-\sqrt{\left(-0.51303\right)^2-4\cdot \:4.905\left(-0.2\right)}}{2\cdot \:4.905}\\\Rightarrow t=0.26088, -0.15629$

Time taken is 0.26088 seconds

$x=u_xt+\dfrac{1}{2}a_xt^2\\\Rightarrow x=1.40953\times 0.26088\\\Rightarrow x=0.3677181864\ m$

The distance the beetle travels on the ground is 0.3677181864 m

6 0

2 years ago

When the mass of the cylinder increased by a factor of 3, from 1.0 kg to 3.0 kg, what happened to the cylinder’s gravitational p

Gnesinka [82]

Answer:

It increased by a factor of 3.

Explanation:

The gravitational potential energy of an object is given by

$U=mgh$

where

m is the mass

g is the gravitational acceleration

h is the heigth of the object relative to some reference point (for instance, the ground)

As we see from the formula, the gravitational potential energy is directly proportional to the mass, m: therefore, if the mass of the cylinder is increased by a factor 3, then the gravitational potential energy will also increase by a factor 3.

6 0

2 years ago