Answer:
1.5 m/s²
Explanation:
Begin by sketching a free body diagram. Three forces are at play on the sea lion: the force of gravity acting downwards, the normal force that is perpendicular to the ramp, and the frictional force parallel to the ramp.
Considering the forces perpendicular to the incline:
∑F = ma
N − mg cos θ = 0
This gives us N = mg cos θ
Next, examining the forces parallel to the incline:
∑F = ma
mg sin θ − Nμ = ma
Substituting for N yields:
mg sin θ − (mg cos θ) μ = ma
g sin θ − g cos θ μ = a
hence a = g (sin θ − μ cos θ)
If we set θ = 23° and μ = 0.26:
a = 9.8 (sin 23 − 0.26 cos 23)
this results in a = 1.48
When rounded to two significant figures, the acceleration of the sea lion is 1.5 m/s².
