Una masa de 10,0 kg se mueve hacia arriba en un plano inclinado con una fuerza de 99,9 N. Si el coeficiente de fricción cinético es de 0,400, ¿Cuál es el trabajo realizado por la fuerza de fricción cinética si la masa se desplaza 15,0 metros hacia arriba?
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Initial velocity,  u = 4.9 m/sfinal velocity = v m/sinitial height,  h₁ =  245 m    final height, h₂ = 0a = -9.8  m/s²time taken = t secondss = ut + 0.5at²⇒  (h₂-h₁) = ut + 0.5at²⇒  0-245 = 4.9t + 0.5×(-9.8)×t²⇒ -245 = 4.9t - 4.9t²⇒ 4.9t² -4.9t -245 =0solving it, we get   t = 7.59sv  = u +  at = 4.9 -9.8×7.59 = 4.9 - 74.38 = -69.48 m/svelocity is  69.48 m/s downward

Iassume the answer is 75 feet. inwards to outwards rather than outwards to inwards. probably be 75.hachures are stokes drawn in direction of the steepest slope. a series of short parallel lines drawn on a map to indicate topographic relief. it is actually called hachures on contour lines. the small marks in the center of the feature are called hachures.observe the picture attached