Question

A spring is stretched 175 mm by an 8-kg block. the block is displaced 100 mm downward from its equilibrium position and given a downward velocity of 1.50 m/s. assume that positive displacement y is downward.

Answer 1

The potential energy stored in a spring is given by the equation PE = 0.5kx². The speed of the block when it crosses the point where the spring is neither compressed nor stretched can be determined using the principle of conservation of energy. The speed of the block when it has traveled a distance of 20 cm from where it was released can be determined by considering the conservation of mechanical energy.

Step-by-step explanation:

The potential energy stored in a spring is given by the equation PE = 0.5kx², where k is the spring constant and x is the displacement from the equilibrium position. In this case, the spring is stretched by 175 mm (or 0.175 m) and the block is displaced 100 mm (or 0.1 m) downward.

Therefore, the potential energy stored in the block-spring system when the block was just released is:

PE = 0.5 * k * (0.175)²

Since the spring constant is not given in the question, we cannot calculate the exact potential energy. To determine the speed of the block when it crosses the point where the spring is neither compressed nor stretched, we need to use the principle of conservation of energy.

When the block is released, the potential energy stored in the spring is converted into kinetic energy:

PE = KE = 0.5mv²

Where m is the mass of the block and v is its velocity.

To determine the speed of the block when it has traveled a distance of 20 cm from where it was released, we need to consider the conservation of mechanical energy.

Answer 2

Let spring K be constant
K= F/x = mg/x = (8kg)(9.81 m/s)/175 × 10⁻³m
= 448.46 N/m
The angular of spring's frequency is
w = √k/√m = √448.46/√8 = 7.49 rad/s

∴ differential equation
y¹¹ + (7.49)²
y = 0
y¹¹ ₊ 56.1
where y = 0.