To solve this problem, we will use the principle of conservation of energy.
Initially, the box has kinetic energy as it moves along the ground. As it slides up the ramp, this kinetic energy is converted into potential energy. When the box reaches the height of 2.5 meters, all the initial kinetic energy will be converted into potential energy, and the box will momentarily stop.
Here's the step-by-step solution to find the initial speed [tex]\( v \)[/tex] required:
1. Write down the expressions for kinetic energy (KE) and potential energy (PE):
- Kinetic Energy (KE) is given by [tex]\( \text{KE} = \frac{1}{2} m v^2 \)[/tex]
- Potential Energy (PE) is given by [tex]\( \text{PE} = m g h \)[/tex], where [tex]\( g \)[/tex] is the acceleration due to gravity, [tex]\( m \)[/tex] is the mass of the box, and [tex]\( h \)[/tex] is the height.
2. Apply the conservation of energy:
- At the beginning, the box has only kinetic energy.
- At the top of the ramp, the box has only potential energy.
[tex]\[
\text{KE}_{\text{initial}} = \text{PE}_{\text{final}}
\][/tex]
3. Substitute the expressions for KE and PE:
[tex]\[
\frac{1}{2} m v^2 = m g h
\][/tex]
4. Simplify the equation (the mass [tex]\( m \)[/tex] cancels out on both sides):
[tex]\[
\frac{1}{2} v^2 = g h
\][/tex]
5. Solve for [tex]\( v \)[/tex]:
[tex]\[
v^2 = 2 g h
\][/tex]
[tex]\[
v = \sqrt{2 g h}
\][/tex]
6. Plug in the known values [tex]\( g = 9.8 \, \text{m/s}^2 \)[/tex] and [tex]\( h = 2.5 \, \text{m} \)[/tex]:
[tex]\[
v = \sqrt{2 \times 9.8 \, \text{m/s}^2 \times 2.5 \, \text{m}}
\][/tex]
7. Calculate the value:
[tex]\[
v = \sqrt{2 \times 9.8 \times 2.5}
\][/tex]
[tex]\[
v = \sqrt{49}
\][/tex]
[tex]\[
v = 7.0 \, \text{m/s}
\][/tex]
Therefore, the initial speed required for the box to slide up the ramp to a height of 2.5 meters is [tex]\( 7.0 \, \text{m/s} \)[/tex].
The answer is [tex]\( \boxed{7.0 \, \text{m/s}} \)[/tex].
