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A=4.2 cm = 0.045 m.
v(max) =Aω=A•sqrt(k/m)=...
A cart of mass 223 g is placed on a frictionless horizontal air track. A spring having a spring constant of 10.20 N/ m is attached between the cart and the left end of the track. When in equilibrium, the cart is located 11.0 cm from the left end of the track. If the cart is displaced 4.20 cm from its equilibrium position, find
(b) its maximum speed
Elena
answered11 years ago
Explain Bot
answered1 year ago
To find the maximum speed of the cart, we can use the principle of conservation of energy.
1. First, let's find the potential energy stored in the spring when the cart is displaced from its equilibrium position. The potential energy stored in a spring is given by the formula:
Potential Energy = 0.5 * k * x^2
Where k is the spring constant and x is the displacement from the equilibrium position.
Plugging in the values:
Potential Energy = 0.5 * 10.20 N/m * (0.0420 m)^2
2. Next, let's use the conservation of energy principle. At maximum speed, all of the potential energy stored in the spring will be converted into kinetic energy. So, the equation becomes:
Potential Energy = Kinetic Energy
3. Kinetic energy is given by the formula:
Kinetic Energy = 0.5 * m * v^2
Where m is the mass of the cart and v is the velocity.
4. Set the potential energy equal to the kinetic energy and solve for v:
0.5 * 10.20 N/m * (0.0420 m)^2 = 0.5 * 0.223 kg * v^2
5. Simplify the equation:
(10.20 N/m * 0.0420 m^2) = (0.223 kg * v^2)
6. Solve for v:
v^2 = (10.20 N/m * 0.0420 m^2) / 0.223 kg
v^2 = 0.1944 m^2/s^2 / 0.223 kg
v^2 ≈ 0.8726 m^2/s^2
7. Take the square root to find v:
v ≈ √(0.8726 m^2/s^2)
v ≈ 0.934 m/s
Therefore, the maximum speed of the cart is approximately 0.934 m/s.