Question

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

kinetic friction between the mass and the incline is .185.
a) What is the value of the of the acceleration of the sled

Answer 1

Answer:
`5.64\ m/s^2`

Step-by-step 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* \cos 45^(\circ))\\a=5.64\ m/s^2`