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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 .

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lublana

Answer:

0.123

Explanation:

We are given that

Constant speed=1.5 m/s

[tex]\theta=22^{\circ}[/tex]

[tex]m=70 kg[/tex]

[tex]F=87 N[/tex]

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

Horizontal pulling force=[tex]F_x=87 cos22^{\circ}=80.649 N[/tex]

Norma force on the sled=[tex]N=mg-87 sin 22^{\circ}[/tex]

N=[tex]70(9.81)-87 sin 22=654.51 N[/tex]

We know that

Friction force=[tex]\mu N[/tex]

[tex]\mu(654.51)=80.649[/tex]

[tex]\mu=\frac{80.649}{654.51}=0.123[/tex]

Hence, the coefficient of friction=0.123

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