Question

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 :
0
(B) 1.71 m
(A) 1.17 m
(C) 7.11 m
(D) 11.7 m​

Answer 1

Answer:

Distance is 1.17 m

Explanation:

from the law of energy conservation; the initial kinetic energy will be distributed among the kinetic energy during movement and potential energy when the block reaches its destination;

Equation would become;  v is the velocity at highest point

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

½ m〖Vo〗^2 = ½ m〖V〗^2 + mgh

m*6^2/2 = m*10*1.1+m*V^2/2

36/2=10*1.1+v2/2

Solving the equation;

v=√14 m/s  

a=−0.6g

=−6 m/s2

from third eqn of motion

S=V^2/2a

=14/12

=1.17 m