Answer:
d= 63.77 m
Explanation:
Newton's second law of the block of ice: Â
∑F = m*a Formula (1)
∑F : algebraic sum of the forces in Newton (N)
m : mass in kilograms (kg)
a : acceleration in meters over second square (m/s²)
Known data
m =125  kg  :mass of the block of ice:
θ = 20°  :angle θ of the slope with respect to the horizontal .
g = 9.8 m/s² : acceleration due to gravity.
Forces acting on the block of ice:
We define the x-axis in the direction parallel to the movement of the block of ice on the  slope  and the y-axis in the direction perpendicular to it.
W: Weight of the box  : In vertical direction
N : Normal force : perpendicular to the direction the slope
Calculated of the weight  of the block of ice
W= m*g Â
x-y weight components
Wx= Wsin θ= ( m*g )*sin(30)°
Wy= Wcos θ =(m*g)*cos(30)°
We apply the formula (1) to calculated acceleration of the block of ice:
∑Fx = m*ax  ,  ax= a  : acceleration of the block  of ice
Wx = m*a
( m*g )*sin(30)° =  m*a  ,We divided both sides of the equation by m
a = g*sin(30) = 9.8*0.5
a = 4.9 m/s²
Kinematics of the block of ice
Because the block of ice moves with uniformly accelerated movement we apply the following formula :
vf²=v₀²+2*a*d  Formula (2
Where: Â
d:displacement (m
vâ‚€: initial speed  (m/s) Â
vf: final speed (m/s) Â
a : acceleration ( m/s² )
Data
vâ‚€ = 0
vf = 25 m/s
a = 4.9 m/s²
We replace data in the formula (2) Â
(vf) ² = v₀²+2*(a)*(d)
(25) ² = 0+2*( 4.9)*(d)
625 = 0+9.8*(d)
d= 625 /9.8
d= 63.77 m
