Abox is sliding along a frictionless surface and gets to a ramp. disregarding friction, how fast should the box be going on the ground in order to slide up the ramp to a height of 2.5 meters, where it stops? (use g = 9.8 m/s2.) a) 2.2 m/s b) 7.0 m/s c) 9.9 m/s d) 24 m/s

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;

Thus, the speed of the box sliding along the frictionless surface is 7 m/s.

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This is amazing.  When you read the quest ion, you wouldn't think there's enough information there to find an answer.  But there is !

-- 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)

v 154.9= 0.9 length length s little over 154.9cm

f=ma newton's second law 7x8=56n