Question

It's a snowy day and you're pulling a friend along a level road on a sled. You've both been taking physics, so she asks what you think the coefficient of friction between the sled and the snow is. You've been walking at a steady 1.5m/s, and the rope pulls up on the sled at a 22 ∘ angle. You estimate that the mass of the sled, with your friend on it, is 70 kg and that you're pulling with a force of 87 N .

Answer 1

0.123

Step-by-step explanation:

We are given that

Constant speed=1.5 m/s

`\theta=22^(\circ)`

`m=70 kg`

`F=87 N`

We have to find the coefficient of friction between sled and the snow.

Horizontal pulling force=
`F_x=87 cos22^(\circ)=80.649 N`

Norma force on the sled=
`N=mg-87 sin 22^(\circ)`

N=
`70(9.81)-87 sin 22=654.51 N`

We know that

Friction force=
`\mu N`

`\mu(654.51)=80.649`

`\mu=(80.649)/(654.51)=0.123`

Hence, the coefficient of friction=0.123