In the figure below, a block slides along a path that is without friction until

Question

In the figure below, a block slides along a path that is without friction until the block reaches the section of length L = 0.95 m, which begins at height h = 2.0 m on a ramp of angle = 30°. In that section, the coefficient of kinetic friction is 0.38. The block passes through point A with a speed of 8.0 m/s. If the block can reach point B (where the friction ends), what is its speed there, and if it cannot, what is its greatest height above A?

My Nonas In the fgure below, a block slides along tn the figurc below, a path that ini tout friction until te block rcachcs the section of ength L 0 DS m which bcgins t hcight h-2.0 m cn a romp of angle 0 30°. In that section, the cocfficicnt of kinetic friction is 0.38. Thc block passes through point A with spcod of 8.0 m/s. If tha blcck can reach point B (wharc the fiction ends), what is its spced there, and if it cannot, what is its greatest hcight abovc A?

Explanation / Answer

here,

height , h = 2 m

theta = 30 degree

initial speed , u = 8 m/s

let the speed before the rough plane be v

using conservation of energy

0.5 * m * (- v^2 + u^2) = m * g * h

0.5 * ( 8^2 - v^2) = 9.81 * 2

v = 4.98 m/s

deaccelration due to friction , a = - ( g * sin(theta) + uk * g * cos(theta))

a = - ( 9.81 * sin(30) + 0.38 * 9.81 * cos(30) )

a = - 8.13 m/s^2

let the speed at B be v'

v'^2 - v^2= 2 * a * L

v'^2 - 4.98^2 = - 2 * 8.13 * 0.95

v' = 3.1 m/s

the speed at point B is 3.1 m/s

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