Alarge block of ice is moving down the hill toward you at 25.0 m/s. its mass is 125 kg. it is sliding down a slope that makes a 30.0 degree angle with the horizontal. in short: think avalanche. assume the block started stationary and moves down the hill with zero friction. how many meters has it been sliding?

d= 63.77 m

Explanation:

Newton's second law of the block of ice:  

∑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  

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:  

d:displacement (m

v₀: initial speed  (m/s)  

vf: final speed (m/s)  

a : acceleration ( m/s² )

Data

v₀ = 0

vf = 25 m/s

a = 4.9 m/s²

We replace data in the formula (2)  

(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

. west

54 degrees incident angle = reflected angle

angle is from normal turned to line of ray