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A ball of mass $m$ is loaded into a launcher with a spring of spring constant $k$ . The ball is pushed down until it is at a vertical position $y=0$ , and the spring is compressed a distance $\Delta L$ , as shown. The ball is then released from rest. Immediately after leaving the launcher, the $x$ - and $y$ -components of the ball's velocity are $v_{x}$ and $v_{y}$ , respectively. The ball reaches a maximum height of $y_{\text {max }}$ and lands a horizontal distance $\Delta x$ away from its initial position. Energy losses due to friction are negligible. Which of the following is a correct conservation of energy equation that compares the total mechanical energy of the ball-spring-Earth system immediately before the ball is launched to the total mechanical energy of the ball-spring-Earth system the moment the ball reaches its maximum height?

Studdy Solution

STEP 1

1. The system consists of a ball, a spring, and the Earth.
2. The spring is compressed by a distance $\Delta L$ .
3. The ball is released from rest.
4. Energy losses due to friction are negligible.
5. We are comparing the total mechanical energy at two points: immediately before the ball is launched and at the maximum height $y_{\text{max}}$ .

STEP 2

1. Identify the forms of energy in the system before the ball is launched.
2. Identify the forms of energy in the system when the ball reaches its maximum height.
3. Write the conservation of energy equation.

STEP 3

Before the ball is launched, the energy in the system is entirely potential energy stored in the spring. The potential energy of the spring is given by:
$E_{\text{spring}} = \frac{1}{2} k (\Delta L)^2$
There is no kinetic energy because the ball is at rest, and the gravitational potential energy is zero because the reference point is $y = 0$ .

STEP 4

At the maximum height $y_{\text{max}}$ , the ball has gravitational potential energy and no kinetic energy because the vertical component of velocity is zero at the maximum height. The gravitational potential energy is given by:
$E_{\text{gravity}} = mgy_{\text{max}}$

STEP 5

According to the conservation of energy, the total mechanical energy before the ball is launched is equal to the total mechanical energy at the maximum height:
$\frac{1}{2} k (\Delta L)^2 = mgy_{\text{max}}$
This equation represents the conservation of energy for the ball-spring-Earth system.
The correct conservation of energy equation is:
$\boxed{\frac{1}{2} k (\Delta L)^2 = mgy_{\text{max}}}$

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