Mary is trying to pull Julie on a sled across a flat snowy field. Mary pulls on the rope attached to the sled. Her pulling force is directed horizontally. Julie weighs 109 pounds. The sled weights 12 pounds. If the coefficient of static friction between the sled runners and the snow is 0.42, how much force must Mary pull with (in lbs) to start moving the sled

F = 498.04 lbs

Explanation: The forces acting on the sled and Julie are show in the figure below. In it, we notice that, for the sled and Julie to go accross the field, they only need force of friction, because, force of friction is a force that resists the relative motion of surfaces.

Force of friction is given by the formula

where

μ is coefficient of friction

is normal force

Normal force is the force the surface exerts on the object. It is always perpendicular and a force of contact.

In the case of the sled, since it is on a horizontal plane, Normal Force has the same magnitude of Gravitational Force. So

Coefficient of friction is how much friction exists between two surfaces.

Rearraging friction force is

Mass for this system is the sum of Julie and the sled, therefore

m = 109 + 12

m = 121 lb

Calculating Friction Force:

498.04 lbs

LBS is a unit of measurement referred as pound by weight.

In conclusion, force Mary needs to start moving the sled is 498.04 lbs

Ke reqd to move from 0 to 50 will be proportional to 50x50 or 2500. this is equal to w.  to change vel from 50 to 150 you need energy prop to 22500 - 2500 = 20,000. this corresponds to 8w.  best answer is a) . work reqd is prop to velocity squared. 

Let n be the height of the tree, in inches. then: n/1500=58/6666n=87000n=1318.18 in, or 109.84 ft☺☺☺☺