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In the figure a block slides along a track from one level to a higher level, by moving through an
intermediate valley. The track is frictionless until the block reaches the higher level. There a frictional
force stops the block in a distance d. The block's initial speed vo is 6m/s. the height difference his
1.1 m and the coefficient of kinetic friction p is 0.6. The value of dis :

(A) 1.71 m
(B) 1.17 m
(C) 7.11 m
(D) 11.7 m​

Respuesta :

Answer:

(B) 1.17 m.

Explanation:

At the highest point, the conservation of energy equation would becomes;

K.E (initial ) = K.E + P.E;

½ m(V)^2 = ½ m(V)^2 + mgh

(m*6^2)/2 = (m * 9.81 * 1.1) + (m*V^2)/2

36/2 = (9.81 * 1.1) + v^2/2

Solving the equation;

v=√14.44 m/s  

= 3.8 m/s

a = −0.6g

= -0.6*9.81

= −5.88 m/s^2

Using equation of motion,

V^2 = U^2 + 2aS

S = V^2/2a

= 14.44/12

= 1.19 m.

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