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 25 ∘ angle. you estimate that the mass of the sled, with your friend on it, is 75 kg and that you're pulling with a force of 79 n .

0.097

Explanation:

The sled is moving at constant velocity. This means that the acceleration is zero: a=0, so according to Newton's second law:

where m is the combined mass of the sled and your friend, the resultant of the forces along the horizontal direction (the direction we are interested in) must be zero:

(1)

There are only two forces acting along the horizontal direction:

is the horizontal force you are applying to pull the sled, with F = 79 N and

is the frictional force, with being the coefficient of friction, m = 75 kg the mass, and g = 9.8 m/s^2 the acceleration due to gravity. The negative sign is due to the fact that the friction is opposite to the pull applied.

Substituting into (1), we can find :

he kicked it 10 feet straight up into the air? i think this works.

a=d/t

a="triangle"v/t

explanation:

a = acceleration