Question

25.
ALL
A block slides along a track from one level to a higher
level by moving through an intermediate valley as
shown in figure. The track is friction less until the
block reaches the higher level. Then there is friction
force which stops the block at a distance d. The
block's initial speed vois 6.0 ms-l; the height
difference h is 1.1 m, and the coefficient of kinetic
friction u is 0.60. The value of d is :-
u=0.60
u=0
MK.a
S
(1) 1,17 m (2) 3 m (3) 2.34 m (4) 6 m

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Answer 1

the answer is 1.17m option no 1

Answer 2

ANSWER:

The distance d at which it stopped is 1.22 m.

EXPLANATION:

   The block sliding in the track will exhibit kinetic energy until its motion gets stopped at a distance d. At that position, the block will also exhibit potential energy. So the initial speed of the block is given as 6 ms-1 and the final velocity will be zero as the block stops at d. The height at which the block stops is at 1.1 m away from the starting height. Let us consider the starting height as 0 for reference. So the height difference will be 1.1 m only. As the block is getting stopped due to frictional force so the coefficient of kinetic friction will exist in this system which is equal to 0.60.     Thus, according to work energy theorem, the work done by the friction to stop the block at d will be equal to the sum of kinetic energy and potential energy at that point d.

Work done =change in kinetic energy+change in potential energy

-coefficient of kinetic friction*mass*acceleration due to gravity*distance=sum of change in kinetic and potential energy

$-0.6 * m * 9.8 * d=\left(\frac{1}{2} * m * v^{2}-\frac{1}{2} * m * u^{2}\right)+\left(m * g * h-m * g * h_{0}\right)$

As we know that final velocity v and initial height h0 is zero, then

$-0.6 * m * 9.8 * d=\left(-\frac{1}{2} * m * u^{2}\right)+(m * g * h)$

The mass is constant, so

$-0.6 * 9.8 * d=\left(-\frac{1}{2} * 6 * 6\right)+(9.8 * 1.1)$

$-5.88 * d=-18+10.78$

d=1.22 m

The distance d at which it stopped is 1.22 m.