A large 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