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-- When the block is sliding along the flat surface, its kinetic energy is (1/2)(Mass·v²).
-- When it's 2.5m up the ramp and stops, its potential energy is (2.5m)·(Mass·g).
-- If there's no friction anywhere, these energies are equal.
(1/2)(Mass·v²) = (2.5m)·(Mass·g)
(v²/2) = (2.5m) · g
v² = 5m · g
v² = 49 m²/s²
v = 7 m/s (B)
The speed of the box sliding along the frictionless surface is 7 m/s.
The given parameters;
The speed at which the box will be moving in order to attain the given height is calculated by applying the principle of conservation of energy as shown below;
[tex]\frac{1}{2}mv^2 = mgh\\\\v^2 = 2gh\\\\\v = \sqrt{2gh} \\\\v = \sqrt{2\times 9.8 \times 2.5} \\\\v = 7 \ m/s[/tex]
Thus, the speed of the box sliding along the frictionless surface is 7 m/s.
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