A6.60-kg block slides with an initial speed of 1.51 m/s up a ramp inclined at an angle of 26.4 ∘ with the horizontal. the coefficient of kinetic friction between the block and the ramp is 0.57.

use energy conservation to find the distance the block slides before coming to

The distance the block slides before coming to rest is 0.12 m.

The given parameters;

Apply the principle of conservation of energy to determine the distance traveled by the block;

Thus, the distance the block slides before coming to rest is 0.12 m.

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s= 0.11 m

Explanation:

Given that

m = 6.6 kg

Initial velocity ,u= 1.51 m/s

Angle ,θ = 26.4°

Coefficient of friction ,μ = 0.57

The final velocity of the block  = 0 m/s

s=Distance cover by block coming to rest

Now by using energy conservation

2 s (μ g cosθ + g sinθ) =  u²

2 x s x 10 (0.57 x cos26.4° + sin26.4°) = 1.51²

Now by solving above the equation

we find ,s= 0.11 m

Therefore the answer will be 0.11 m.

here we know that ball will drop down by total distance 3.2 m

so here we know that

now using kinematics

now in the same time it will cover horizontal distance of 0.75 m

so speed of the ball will be