Question

A 50 kg skier starting from rest wants to jump a ravine. He begins 19 m from the edge of the cliff, on a 30 degree down slope (with the horizontal). If the lip on the far side of the ravine is 14.26 m below the upper edge, and the ravine width is 10.35 m, with how much constant force would he need to push to get up to the speed needed to make it across?

Also note the coefficient of Kinetic friction is 0.35 this problem is due on Friday 1/25/19 and I really need help, Please show work so I can also understand because i'm kinda lost. Thanks!

Answer 1

let the initial speed of jump be v_o

consider the motion in horizontal direction

X = horizontal displacement =10.35 m

v_oy = initial velocity in horizontal direction = v_o Cos30

t = time of travel

a_x = acceleration = 0 m/s²

using the kinematics equation

X = v_oy t + (0.5) a_x t²

10.35 = (v_o Cos30) t + (0.5) (0)t²

t = 10.35 /(v_o Cos30)                                                          eq-1

consider the motion along the vertical direction :

Y = vertical displacement = - 14.26 m

a_y = acceleration = - 9.8 m/s²

v_oy = initial velocity in vertical direction = v_o Sin30

t = time of travel = ?

using the kinematics equation

Y = v_oy t + (0.5) a_y t²

- 14.26= (v_o Cos30) (10.35 /(v_o Cos30) ) + (0.5) (- 9.8) (10.35 /(v_o Cos30) )²

v_o = 5.33 m/s

d = distance moved before making the jump = 19 m

v_i = initial velocity when it start to move = 0 m/s

a = acceleration

using the equation

W = change in kinetic energy

W = (0.5) m (v_o² - v_i²)

W = (0.5) (50) ((5.33)² - (0)²)

F d = (0.5) (50) ((5.33)² - (0)²)

F (19)= (0.5) (50) ((5.33)² - (0)²)

F = 37.4 N