Work and Energy

Question

A sled is given an initial speed of 15 𝑚/𝑠 up a snow-covered hill. The hill makes an angle of 30° with respect to horizontal, and the coefficient of kinetic friction between the snow and the sled is 0.22. Determine how far up the hill the sled slides before coming to stop.

Answer 1

As we know:

∆K = W_f + W_mg → W_f = - μ_k (mg×cos(θ)).X , W_mg = - (mg).h , h = X. sin(θ) → W_mg = - (mg×sin(θ)). X

h is the height at which the sled will go up vertically, and X is the distance in which the sled is moving on the hill.

W_mg is negative because mg is downward, and sled go up vertically (h). So force and displacement have opposite directions.

W_f is negative because f_k is toward of down of the hill but X goes toward of up of the hill.

Initial velocity is 15 (m/s) and final velocity is zero , ∆K = K₂ - K₁

∆K = W_f + W_mg → (m/2)(0)² - (m/2).(15)² = - μ_k (mg×cos(θ)).X - (mg×sin(θ)). X

- (m/2).(15)² = - μ_k (mg×cos(θ)).X - (mg×sin(θ)). X → X = (15)²/ {[μ_k ×cos(θ)+ sin(θ)].g}

Answer 2