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 where due to friction, the block stops in a distance d. The block's initial speed $v_0$ is 6m/s, the height difference h is 1.1 m the coefficient of kinetic friction is 0.6. The value of d is :

• A. 1.17 m
• B. 1.71 m
• C. 7.11m
• D. 11.7m

Answer 1

The correct option is A 1.17 m

Initial velocity of block is 6 m/s

Initial kinetic energy of block is

$K.E.=\frac{1}{2}m{6}^2=18m$

when it reach at the higher level then loss in kinetic energy is equal to gain in potential energy

$h=1.1m$

$P.E.=mghm\times 10\times 1.1=11m$

thus remaining kinetic energy is

18m - 11m = 7m

then velocity at higher starting point is

$\frac{1}{2}m{v}^2=7m$

$v=\sqrt{14}$

if this is assumed as initial velocity

and final velocity will be zero due to retardation of friction

$a=-\mu g=0.6\times 10=-6m/s^2$

by using newtons law of motion

$v^2=u^2+2asv=0u=\sqrt{14}d=\frac{14}{2\times 6}=1.16667m$