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 1.5\;{\rm m}/{\rm s}, and the rope pulls up on the sled at a 42.0^\circ angle. You estimate that the mass of the sled, with your friend on it, is 73.0 kg and that you're pulling with a force of 87.0 N.?

0.0984

Explanation:

From the first diagram attached below; a free flow diagram shows the interpretation of this question which will be used  to solve this question.

From the diagram, the horizontal component of the force is:

Replacing 42°  for θ and 87.0° for

On the other hand, the vertical component  is ;

Replacing 42°  for θ and 87.0° for

However, resolving the vector, let A be the be the component of the mutually perpendicular directions.

The magnitude of the two components is shown in the second attached diagram below and is now be written as A cos θ and A sin θ

The expression for the frictional force is expressed as follows:

Where;

is said to be the coefficient of the friction

N = the  normal force

Similarly the normal reaction (N) = mg - F sin θ

Replacing . The normal reaction can now be:

By balancing the forces, the horizontal component of the force equals to frictional force.

The horizontal component of the force is given as follows:

Making the subject of the formular in the above equation; we have the following:

Replacing the following values: i.e

m = 73 Kh

g  = 9.8 m/s²

Then:

Thus, the coefficient of friction is = 0.0984

not moving. pos accel,