Question

It's a snowy day and you're pulling a friend along a

Answer 1

The coefficient of friction between the sled and the snow is 0.119.

To find the answer, we need to know about the friction.

How to find the coefficient of friction between the sled and the snow?

• Whenever a body moves over the surface of another body, a force come into play, which acts parallel to the surface of contact and oppose the relative motion. This opposing force is called friction.
• To solve the problem, we have to draw the free body diagram of the given system.
• We have given with the following values,

                                     $a=0\\\alpha =35^0\\T=75N\\m=57kg$

Here, acceleration will be equal to zero, because the velocity is given as constant.

• Thus, from the diagram, we can write the balancing equations as follows,

                                      $ma=Tcos\alpha -f\\\where\\f=kN\\\N+Tsin\alpha=mg\\Thus,\\N=mg-Tsin\alpha$

• Substituting N in f and f in the equation of ma, then we get,

                   $ma= Tcos\alpha -k(mg-Tsin\alpha )$

• Substituting values, we get the coefficient of friction as,

                    $0=(75*cos35)-k((57*9.8)-(75sin35))\\\\k((57*9.8)-(75sin35))=(75*cos35)\\\\515.6k=61.44\\\\k=\frac{61.44}{515.6}=0.119$

Thus, we can conclude that, the coefficient of friction between the sled and the snow is 0.119.

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Answer 2

The sled's coefficient of friction with the snow is 0.119.

We must understand the friction in order to choose the solution.

How can I determine the sled and snow's coefficient of friction?

• A force that works parallel to the surface of contact and opposes the relative motion is present whenever one body moves over the surface of another body. Friction is the name for this opposing force.

• We must create the given system's free body diagram in order to solve the issue.

• The values that we have provided are

                               $\alpha =35\\T=75N\\m=57kg\\a=0$

Because the velocity is specified as constant in this case, the acceleration will be equal to zero.

• Consequently, we can express the balancing equations as follows using the diagram:

                             $ma=Tcos\alpha -f\\ where,f=kN\\N+Tsin\alpha =mg\\ thus,\\N=mg-Tsin\alpha$

• When we substitute N for f and f in the equation for ma, we obtain,

                       $ma=Tcos\alpha -k(mg-Tsin\alpha )$

• By substituting values, we obtain the friction coefficient as.

                                   $k=0.119$

As a result, we may say that there is 0.119 coefficient of friction between the sled and the snow.

Learn more about the friction here:

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