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.

Learn more here:link

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.

the answer is f1 and d2 because if you put equal force on those two points the beam would balance trust me on this