Question

4. A 5.00-kg sled starts at rest and accelerates down a 45.0° incline. The coefficient of

Answer 1

Answer: $5.64\ m/s^2$

Explanation:

Given

mass of sled m=5 kg

Elevation $\theta =45^{\circ}$

$\mu _k=0.185$ (coefficient of kinetic friction)

Friction will oppose the motion of the sled while sled weight sin component helps it to move down the incline

$W\sin\theta-F_r=ma\\where\\W=weight(mg)\\F_r=Friction(\mu_kmg\cos\theta)\\a=acceleration$

Substituting values

$mg\sin \theta-\mu_kmgcos\theta=ma\\g\sin \theta-\mu_kg\cos \theta=a\\a=g(\sin 45^{\circ}-0.185\times \cos 45^{\circ})\\a=5.64\ m/s^2$