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Triss [41]
3 years ago
14

The spring of modulus k = 200 n /m is compressed a distance of 300 mm and suddenly released with the system at rest. determine t

he absolute velocities of both masses when the spring is unstretched. neglect friction.

Physics
1 answer:
DiKsa [7]3 years ago
8 0
I attached the missing picture.
Let's analyze the situation as spring goes from stretched to unstretched state.
When you stretch the string you have to use force against ( you are doing work) this energy is then stored in the spring in the form of potential energy. When we release the spring the energy is being used to push the two carts. When the spring reaches its unstretched length its whole initial potential energy has been used on the carts, and this is the moment when two carts have maximum velocity.
The potential energy of compressed ( stretched) spring is:
E_p=\frac{1}{2}kx^2
The kinetic energy of two carts is:
E_{k1}+E_{k2}=m_1\frac{v_1^2}{2}+m_2\frac{v_2^2}{2}
So we have:
E_p=E_{k1}+E_{k2}\\ \frac{1}{2}kx^2=m_1\frac{v_1^2}{2}+m_2\frac{v_2^2}{2}
Momentum also has to be conserved:
m_1v_1-m_2v_2=0\\ m_1v_1=m_2v_2\\ v_1=\frac{m_2}{m_1}v_2
Momentum before the release of the spring is zero so it has to stay zero. We plug this back into the expresion we got from law of conservation of energy and we get:
v_2^2=\frac{m_1^2}{m_2^2-m_1^2}kx^2=4.05\\
v_2=\sqrt{4.05}=2.012\frac{m}{s}
Now we go back to the momentum equation:
v_1=\frac{m_2}{m_1}v_2\\
v_1=4.69\frac{m}{s}

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When turned on, a fan requires 5.0 seconds to get up to its final operating rotational speed of 1200 rpm. a) How large is the fi
vova2212 [387]

Answer:

a)

125.6 rad/s

b)

25.12 rad/s²

Explanation:

a)

t = time required by the fan to get up to final operating speed = 5 sec

w = final operating rotational speed = 1200 rpm

we know that :

1 revolution = 2π rad

1 min = 60 sec

w = 1200\frac{rev}{min}\frac{2\pi rad}{1 rev}\frac{1 min}{60 sec}

w = \frac{1200\times 2\pi }{60}\frac{rad}{s}

w = 125.6 rad/s

b)

w₀ = initial angular speed = 0 rad/s

α = angular acceleration

using the equation

w = w₀ + α t

125.6 = 0 + α (5)

α = 25.12 rad/s²

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3 years ago
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3 years ago
A net force of 345 N accelerates a boy on a sled at 3.2 m/s^2 . What is combined mass of the sled
Daniel [21]

Answer:

Mass, m = 26.54kg

Explanation:

Net force can be defined as the vector sum of all the forces acting on a body or an object i.e the sum of all forces acting simultaneously on a body or an object.

Mathematically, net force is given by the formula;

Fnet = Fapp + Fg

Where;

  • Fnet is the net force
  • Fapp is the applied force
  • Fg is the force due to gravitation

<u>Given the following data;</u>

Net force, Fnet = 345

Acceleration, a = 3.2m/s²

<u>To find mass;</u>

Fnet = Fapp + Fg

Fnet = ma + mg

Fnet = m(a+g)

m = Fnet/(a+g)

We know that acceleration due to gravity, g = 9.8m/s²

Substituting into the equation, we have;

m = 345/(3.2 + 9.8)

m = 345/13

Mass, m = 26.54kg

6 0
2 years ago
A driver who does not wear a seat belt continues to move at the initial velocity until she or he hits something solid (e.g the s
egoroff_w [7]

This question is incomplete, the complete question is;

Seatbelts provide two main advantages in a car accident (1) they keep you from being thrown from the car and (2) they reduce the force that acts on your during the collision to survivable levels. This second benefit can be illustrated by comparing the net force encountered by a driver in a head-on collision with and without a seat beat.  

1) A driver wearing a seat beat decelerates at roughly the same rate as the car it self. Since many modern cars have a "crumble zone" built into the front of the car, let us assume that the car decelerates of a distance of 1.1 m. What is the net force acting on a 70 kg driver who is driving at 18 m/sec and comes to rest in this distance?

Fwith belt =

2) A driver who does not wear a seat belt continues to move at the initial velocity until she or he hits something solid (e.g the steering wheel) and then comes to rest in a very short distance. Find the net force on a driver without seat belts who comes to rest in 1.1 cm.

Fwithout belt =

Answer:

1) The Net force on the driver with seat belt is 10.3 KN

2) the Net force on the driver without seat belts who comes to rest in 1.1 cm is 1030.9 KN

Explanation:

Given the data in the question;

from the equation of motion, v² = u² + 2as

we solve for a

a = (v² - u²)/2s ----- let this be equation 1

we know that, F = ma ------- let this be equation 2

so from equation 1 and 2

F = m( (v² - u²)/2s )

where m is mass, a is acceleration, u is initial velocity, v is final velocity and s is the displacement.

1)

Wearing sit belt, car decelerates of a distance of 1.1 m. What is the net force acting on a 70 kg driver who is driving at 18 m/sec and comes to rest in this distance.

i.e, m = 70 kg, u = 18 m/s, v = 0 { since it came to rest }, s = 1.1 m

so we substitute the given values into the equation;

F = 70( ((0)² - (18)²) / 2 × 1.1 )

F = 70 × ( -324 / 2.4 )

F = 70 × -147.2727

F = -10309.09 N

F = -10.3 KN

The negative sign indicates that the direction of the force is opposite compared to the direction of the motion.

Fwith belt =  10.3 KN

Therefore, Net force of the driver is 10.3 KN

2)

No sit belt,  

m = 70 kg, u = 18 m/s, v = 0 { since it came to rest }, s = 1.1 cm = 1.1 × 10⁻² m

we substitute

F = 70( ((0)² - (18)²) / 2 × 1.1 × 10⁻² )

F = 70 × ( -324 / 0.022 )

F = 70 × -14727.2727

F = -1030909.08 N

F = -1030.9 KN

The negative sign indicates that the direction of the force is opposite compared to the direction of the motion.

Fwithout belt = 1030.9 KN

Therefore, the net force on the driver without seat belts who comes to rest in 1.1 cm is 1030.9 KN

4 0
3 years ago
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