Question

A 2.0-kg block sliding on a rough surface is attached to one end of a horizontal spring (k = 200 N/m) which has its other end fixed. If the block has a speed of 4.0 m/s as it passes through the equilibrium position, what is its speed when it is 10 cm from the equilibrium position? The coefficient of friction between the block and surface is 0.15.

Answer 1

To find the speed of the block when it is 10 cm from the equilibrium position, we can use the principles of conservation of energy and Hooke's Law. By equating the initial kinetic energy to the potential energy at 10 cm from equilibrium, we can solve for the new velocity of the block, which is approximately 1.79 m/s.

Step-by-step explanation:

To find the speed of the block when it is 10 cm from the equilibrium position, we can use the principles of conservation of energy and Hooke's Law.

First, let's consider the initial situation when the block passes through the equilibrium position. At this point, the block possesses only kinetic energy. Using the equation for kinetic energy, we can find the initial velocity of the block:

Kinetic Energy = 1/2 * mass * velocity^2

Since the velocity is 4.0 m/s, we can solve for the initial kinetic energy.

Once we have the initial kinetic energy, we can use the principle of conservation of energy to find the potential energy when the block is 10 cm from the equilibrium position. The potential energy stored in the spring is given by the equation:

Potential Energy = 1/2 * spring constant * displacement^2

Since the displacement is 10 cm, we can solve for the potential energy.

Finally, we can equate the initial kinetic energy to the potential energy at 10 cm from equilibrium to find the new velocity of the block:

1/2 * mass * velocity^2 = 1/2 * spring constant * displacement^2

Solving for the new velocity, we can find that it is approximately 1.79 m/s.

Answer 2

The speed is 3.87 m/s

Step-by-step explanation:

The total work is equal to the change to kinetic energy:

Wtotal = ΔEk

Wspring + Wfriction = Ekf - Eki

Eki = 0

`-((1)/(2) kd^(2) )-(fk*m*g*d^(2) )=(1)/(2) m(v^(2) -v_(i) ^(2) )`

Where

k = 200 N/m

d = 10 cm = 0.1 m

fk = 0.15

m = 2 kg

g = 9.8 m/s²

vi = 4 m/s

Replacing and clearing v:

`-((1)/(2) *200*0.1^(2) )-(0.15*2*9.8*0.1^(2) )=(1)/(2) *2*(v^(2) -4^(2) )\\-1-0.0294=v^(2) -16\\v=√(-1-0.0294+16) =3.87m/s`