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Sagot :
To determine the speed the box must have on the ground in order to slide up the ramp to a height of 2.5 meters, we can use the principle of conservation of energy. Essentially, the kinetic energy on the ground must equal the potential energy at the top of the ramp.
1. Potential Energy (PE) at the top of the ramp:
At the top of the ramp, all the kinetic energy would have been converted into potential energy. The potential energy at height [tex]\( h \)[/tex] is given by the formula:
[tex]\[ PE = mgh \][/tex]
Where:
- [tex]\( m \)[/tex] is the mass of the box (in kilograms).
- [tex]\( g \)[/tex] is the acceleration due to gravity ([tex]\( 9.8 \, m/s^2 \)[/tex]).
- [tex]\( h \)[/tex] is the height (2.5 meters).
2. Kinetic Energy (KE) on the ground:
On the ground, the box’s energy is entirely kinetic, given by:
[tex]\[ KE = \frac{1}{2} mv^2 \][/tex]
Where:
- [tex]\( v \)[/tex] is the velocity we need to find.
3. Energy Conservation:
At the point of takeoff (on the ground), kinetic energy converts fully to potential energy at the ramp's summit. Therefore:
[tex]\[ \frac{1}{2} mv^2 = mgh \][/tex]
Notice that [tex]\( m \)[/tex] (mass of the box) cancels out:
[tex]\[ \frac{1}{2} v^2 = gh \][/tex]
4. Solve for [tex]\( v \)[/tex]:
[tex]\[ v^2 = 2gh \][/tex]
[tex]\[ v = \sqrt{2gh} \][/tex]
5. Substitute the known values:
[tex]\[ v = \sqrt{2 \times 9.8 \, m/s^2 \times 2.5 \, \text{meters}} \][/tex]
6. Perform the calculation:
[tex]\[ v = \sqrt{49} \][/tex]
[tex]\[ v = 7.0 \, m/s \][/tex]
Therefore, the velocity required for the box to reach a height of 2.5 meters on the ramp is [tex]\( 7.0 \, m/s \)[/tex].
Among the given choices:
- [tex]\( 2.2 \, m/s \)[/tex]
- [tex]\( 7.0 \, m/s \)[/tex]
- [tex]\( 9.9 \, m/s \)[/tex]
- [tex]\( 24 \, m/s \)[/tex]
The correct answer is [tex]\( 7.0 \, m/s \)[/tex].
1. Potential Energy (PE) at the top of the ramp:
At the top of the ramp, all the kinetic energy would have been converted into potential energy. The potential energy at height [tex]\( h \)[/tex] is given by the formula:
[tex]\[ PE = mgh \][/tex]
Where:
- [tex]\( m \)[/tex] is the mass of the box (in kilograms).
- [tex]\( g \)[/tex] is the acceleration due to gravity ([tex]\( 9.8 \, m/s^2 \)[/tex]).
- [tex]\( h \)[/tex] is the height (2.5 meters).
2. Kinetic Energy (KE) on the ground:
On the ground, the box’s energy is entirely kinetic, given by:
[tex]\[ KE = \frac{1}{2} mv^2 \][/tex]
Where:
- [tex]\( v \)[/tex] is the velocity we need to find.
3. Energy Conservation:
At the point of takeoff (on the ground), kinetic energy converts fully to potential energy at the ramp's summit. Therefore:
[tex]\[ \frac{1}{2} mv^2 = mgh \][/tex]
Notice that [tex]\( m \)[/tex] (mass of the box) cancels out:
[tex]\[ \frac{1}{2} v^2 = gh \][/tex]
4. Solve for [tex]\( v \)[/tex]:
[tex]\[ v^2 = 2gh \][/tex]
[tex]\[ v = \sqrt{2gh} \][/tex]
5. Substitute the known values:
[tex]\[ v = \sqrt{2 \times 9.8 \, m/s^2 \times 2.5 \, \text{meters}} \][/tex]
6. Perform the calculation:
[tex]\[ v = \sqrt{49} \][/tex]
[tex]\[ v = 7.0 \, m/s \][/tex]
Therefore, the velocity required for the box to reach a height of 2.5 meters on the ramp is [tex]\( 7.0 \, m/s \)[/tex].
Among the given choices:
- [tex]\( 2.2 \, m/s \)[/tex]
- [tex]\( 7.0 \, m/s \)[/tex]
- [tex]\( 9.9 \, m/s \)[/tex]
- [tex]\( 24 \, m/s \)[/tex]
The correct answer is [tex]\( 7.0 \, m/s \)[/tex].
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