An ideal spring with a stiffness of 329 N/m is attached to a wall, and its other end is attached to a block that has a mass of 15.0 kg. The spring/block system is then stretched away from the spring's relaxed position until 57.0 J of mechanical energy is stored in the spring. Then the system is released to oscillate freely. With the block sliding horizontally on the level, frictionless floor. Find the speed of the block when the spring is compressed by 18.4 cm.
The maximum deflection A is given by
57 J = (1/2)kA^2.
A is 0.589 m
When the deflection is 0.184 m, the spring's potential energy is
(1/2) k (0.184)^2
Subtract that from the total energy (57J)to get the kinetic energy at that compression, and use that to get the speed.
do it on ur own stop asking for help
ask esgar for help
To find the speed of the block when the spring is compressed by 18.4 cm, we can use the principle of conservation of mechanical energy.
Let's break down the problem into two parts: the potential energy stored in the spring and the kinetic energy of the block.
1. Potential Energy stored in the spring:
The potential energy stored in a spring is given by the formula:
PE = (1/2)kx^2
Where PE is the potential energy, k is the stiffness of the spring, and x is the displacement from the spring's relaxed position.
In this case, the potential energy stored in the spring is 57.0 J, and the stiffness of the spring is 329 N/m. We can use this information to find the value of x.
57.0 J = (1/2)(329 N/m)(x^2)
Multiplying both sides by 2 and dividing by 329 N/m:
x^2 = (2 * 57.0 J) / 329 N/m
x^2 = 0.3463
Taking the square root of both sides, we find:
x ≈ 0.588 m
2. Kinetic Energy of the block:
The kinetic energy of an object is given by the formula:
KE = (1/2)mv^2
Where KE is the kinetic energy, m is the mass of the object, and v is the velocity of the object.
We know the mass of the block is 15.0 kg. To find the velocity of the block when the spring is compressed by 18.4 cm, we need to use the law of energy conservation.
When the spring is compressed by 18.4 cm, all the potential energy is converted into kinetic energy.
PE = KE
(1/2)kx^2 = (1/2)mv^2
Plugging in the values we know:
(1/2)(329 N/m)(0.184 m)^2 = (1/2)(15.0 kg)(v^2)
Multiplying and dividing by 2:
(329 N/m)(0.184 m)^2 = (15.0 kg)(v^2)
Simplifying:
v^2 = [(329 N/m)(0.184 m)^2] / (15.0 kg)
v^2 = 0.15705 m^2/s^2
Taking the square root of both sides, we find:
v ≈ 0.396 m/s
Therefore, the speed of the block when the spring is compressed by 18.4 cm is approximately 0.396 m/s.