A 2-kg block slides on a frictionless horizontal surface and is connected on one side to a spring (k = 40 N/m) as shown in the figure. The other side is connected to a 4-kg block that hangs vertically. The system starts from rest with the spring unextended.
What is the speed of the 4-kg block when the extension is 50 cm?
To find the speed of the 4-kg block when the extension of the spring is 50 cm, we will use the principles of energy conservation.
First, let's find the elastic potential energy stored in the spring when it is extended by 50 cm. We can use the formula for elastic potential energy:
Elastic Potential Energy = (1/2) * k * x^2
where k is the spring constant and x is the displacement from the equilibrium position.
Given that the spring constant (k) is 40 N/m and the displacement (x) is 50 cm (which is 0.5 m), we can substitute these values into the formula:
Elastic Potential Energy = (1/2) * 40 N/m * (0.5 m)^2
= 0.5 * 40 N/m * 0.25 m^2
= 5 Joules
Since the system starts from rest, this elastic potential energy will be converted into kinetic energy of the 4-kg block when the system reaches equilibrium.
Kinetic energy is given by the formula:
Kinetic Energy = (1/2) * m * v^2
where m is the mass of the object and v is its velocity.
In this case, the mass (m) is 4 kg, and we'll find v, the velocity of the 4-kg block when it has traveled the distance of 50 cm from the equilibrium position.
Equating the elastic potential energy to kinetic energy, we have:
5 Joules = (1/2) * 4 kg * v^2
Now we can solve for v:
v^2 = (2 * 5 J) / 4 kg
v^2 = 10 J / 4 kg
v^2 = 2.5 m^2/s^2
Taking the square root of both sides of the equation:
v = √(2.5 m^2/s^2)
v ≈ 1.58 m/s
Therefore, the speed of the 4-kg block when the spring extension is 50 cm is approximately 1.58 m/s.