Question

A sled is being held at rest on a slope that makes an angle θ with the horizontal. After the sled is released, it slides a distance d1 down the slope and then covers the distance d2 along the horizontal terrain before stopping. Find the coefficient of kinetic friction μk between the sled and the ground, assuming that it is constant throughout the trip.

Answer 1

Answer:
`\mu_(k)=(a)/(g) csc \theta+ ctan \theta`

Step-by-step explanation:

If we draw a free body diagram of the sled we will have:

`\sum{F_(x)}=F + W_(x) - F_(friction)=0`

Where:

`F=m.a` being
`m` the mass of the car and
`a` its acceleration.

`W_(x)=Wcos\theta` is the horizontal component of the weight of the car

`F_(friction)=\mu_(k)N` is the friction force where
`N` is the normal force

Then:

`m.a + Wcos\theta - \mu_(k)N=0` (1)

`\sum{F_(y)}=N-W_(y)=0`

Where
`W_(y)=Wsin \theta` is the vertical component of the weight, then:

`N=Wsin \theta` (2)

Substituting (2) in (1):

`m.a + Wcos\theta - \mu_(k)Wsin \theta=0` (3)

Finding
`\mu_(k)`:

`\mu_(k)=(m.a + Wcos \theta)/(Wsin \theta)` (4)

Simplifying:

`\mu_(k)=(a)/(g) csc \theta+ ctan \theta`