This athlete loads a sled to a combined mass of 90.0 kg, and drags the sled with a force of 258 N at an angle of 15.0° across a surface with a coefficient of kinetic friction of 0.250. What is the approximate acceleration of the sled? Group of answer choices

The approximate acceleration of the sled is 0.319 m/s²

Explanation:

Given;

mass of the sled, m = 90 kg

horizontal applied force, f = 258 N

angle of inclination of the force, θ = 15°

coefficient of kinetic friction of 0.250, μ = 0.25

The net vertical force on the sled is given by;

Fcosθ - μmg = ma

258Cos15 - (0.25 x 90 x 9.8) = ma

249.2 - 220.5 = ma

28.7 = ma

a = 28.7 /m

a = 28.7 / 90

a = 0.319 m/s²

Therefore, the approximate acceleration of the sled is 0.319 m/s²

force is mass*acceleration.

we have a very small mass: 11.4*0.001 kg

if we want 0.035n of force, we need an acceleration:

0.035 = 11.4*0.001*a

a = 0.035/(11.4*0.001) m/s^2

nah.   i'm just sleepy.