A block of mass 0.45 kg can slide over a frictionless horizontal surface. It is attached to a spring whose stiffness constant is k = 18 N/m. The block is pulled 38 cm and let go.
a)What is its maximum speed?
b)What is its speed when the extension is 30 cm?
c)At what position is the kinetic energy equal to the potential energy?
To solve this problem, we will use the concepts of energy and Hooke's law, which relates the force exerted by a spring to its extension.
a) To find the maximum speed of the block, we need to determine the maximum potential energy stored in the spring, which is converted into kinetic energy at that point.
The potential energy stored in a spring is given by the equation:
Potential Energy (PE) = (1/2)kx^2
where k is the stiffness constant of the spring (given as 18 N/m) and x is the displacement (given as 38 cm = 0.38 m).
PE_max = (1/2)(18 N/m)(0.38 m)^2
PE_max = 0.6516 J
At the maximum speed, all the potential energy is converted to kinetic energy.
Kinetic Energy (KE) = (1/2)mv^2
where m is the mass of the block (given as 0.45 kg) and v is the maximum speed.
Setting the potential energy equal to the kinetic energy, we have:
0.6516 J = (1/2)(0.45 kg)v^2
Now, we can solve for v:
v^2 = (2 * 0.6516 J) / (0.45 kg)
v^2 = 2.892 J/kg
v = √(2.892 J/kg)
v ≈ 1.70 m/s
Therefore, the maximum speed of the block is approximately 1.70 m/s.
b) To find the speed when the extension is 30 cm (0.30 m), we can use the same formula for kinetic energy.
PE = (1/2)kx^2
0.6516 J = (1/2)(18 N/m)(0.30 m)^2
Now we can solve for the kinetic energy at this point:
KE = 0.6516 J
Using the formula for kinetic energy:
KE = (1/2)mv^2
0.6516 J = (1/2)(0.45 kg)v^2
Now, solve for v:
v^2 = (2 * 0.6516 J) / (0.45 kg)
v^2 = 2.892 J/kg
v = √(2.892 J/kg)
v ≈ 1.70 m/s
Therefore, the speed of the block when the extension is 30 cm is approximately 1.70 m/s.
c) To find the position where the kinetic energy is equal to the potential energy, we can set the two equations equal to each other:
(1/2)kx^2 = (1/2)mv^2
The mass of the block and the stiffness constant of the spring are given, so we can substitute these values in:
(1/2)(18 N/m)x^2 = (1/2)(0.45 kg)v^2
Now, substitute the values we found in parts (a) and (b):
(1/2)(18 N/m)x^2 = (1/2)(0.45 kg)(1.70 m/s)^2
Now we can solve for x:
x^2 = (0.45 kg)(1.70 m/s)^2 / (18 N/m)
x^2 = 0.2067 m^2
x = √(0.2067 m^2)
x ≈ 0.454 m
Therefore, the position where the kinetic energy is equal to the potential energy is approximately 0.454 m.