The only trick here is to break up the force into horizontal components. The pushing force is the horizontal component. I assume you can do that. The vertical component of the pushing force does no work, no does gravity, as there is no movement in the vertical direction.
On the second, I would break it into parts, down the hill, and horizontal.
Potential energy at top of hill=fricion work downhill + friction work horizontal.
PE=mu*mass*g*cosTheta*150 + mu*mass*g*distance.
Notice mass will divide out when you put mgh in for PE, so then solve for distance.
A block of mass 2.50 kg is pushed 2.40 m along a frictionless horizontal table by a constant 10.0 N force directed 25.0° below the horizontal.
(a) Determine the work done by the applied force.
J
(b) Determine the work done by the normal force exerted by the table.
J
(c) Determine the work done by the force of gravity.
J
(d) Determine the work done by the net force on the block.
J
2. A skier starts from rest at the top of a hill that is inclined at 9.5° with the horizontal. The hillside is 150 m long, and the coefficient of friction between snow and skis is 0.0750. At the bottom of the hill, the snow is level and the coefficient of friction is unchanged. How far does the skier glide along the horizontal portion of the snow before coming to rest?
m
3 answers
im still really confused
work is done when a force operates in the direction of the force.
work=force*distance, when the force and distance are in the same direction.
work=force*distance, when the force and distance are in the same direction.
1 answer