please help! im so stressed with these 3 problems?

Question

1. A worker pulls a sled with a force of 80 N directed at an angle of 60 degrees above the horizontal over a level distance of 6 m. If a frictional force of 24 N acts on the sled in a direction opposite to that of the worker, what net work is done on the sled?

a. 240 J, b. 216 J, c. 144 J, d. 96 J


2. A model rocket sits on the launch pad until its fuel is ignited, blasting the rocket upward. During the short time of blast-off, as the ignited fuel goes down, the rocket goes up because:

a. momentum is conserved in this process
b. the fuel pushes on the ground
c. air friction pushes on the escaping fuel
d. the downward force of gravity is less than the downward momentum of the fuel

please explain why you believe the answers are what they are just in case yours are different than mine. thanks a lot guys!

3. A 10 g bullet with an initial speed of 100 m/s is fired horizontally into a 90 g wodden block initially at rest on a frictionless horizontal surface. The bullet passes completely through the block and emerges with a final speed of 20 m/s in the horizontal direction. How much mechanical energy is lost in this process?

a. 44.4 J
b. 50 J
c. 26.4 J
d. 20 J

Answer 1

1- d 96 j

work is done by a force applying on a distance, the fraction of the force that would pull vertically on the sled does not produce any work if the sled stays on the groud. So in the end, the force overcame the resistance of 24 N over a distance of 6 m, which is (24 * 6) =144 J absorbed by friction, and cos(60)*80 N = 40 N * 6 = 240 J expanded; with a net result of 240 - 144 = 96 J being put as useful work accelerating the sled.

2- d

it is a question of action=reaction, and the force of gravity has to be overcome by the sheer momentum of the expanding exhaust gasses.

3- a 44.4 J.

There are two things at play here: conservation of momentum, and conservation of energy, a portion of the energy being used to poke a hole in the block of wood., which is the enrgy lost.
The bullet has an emergy of

1/2 m v^2

1/2 .01 100^2 = 50 J

and momentum of .01 * 100 = 1 kg-m/s

After the bullet emerges from the block it is travelling at 20 m/s,
so its energy is 1/2 .01 20^2 = 2 J, and its momentum is .01 * 20 = 0.2 kg-m/s. So the other 0.8 kg-m/s must be in the moving block, and since it has a mass of .09 kg, is must then move at 8.888 m/s, and thus its kinetic energy must be

1/2 .09 8.888^2 = 3.555 J

Since the bullet stil has 2 J left, and the block captured 3.555 J,
the energy f the bullet-block system is 5.555 J, and thus (50 - 5.555) J = 44.444 J was absorbed by the block for the formation of the bullet hole.

Answer 2

1) Horizontal component of man's force
= 80 x cos 60 = 80 x 0.5 = 40 N
Net force 40 - 24 = 16 N
Work done = Force x Distance =16 x 6 = 96 J

2) The correct answer is (a), Momentum concerved.
The momentum of the ignited fuel going downwords is equal to the momentum of the rocket going upwards.

3) Conservation of momentum
10 x 100 = 10x 20 + 90 x V, where V= velocity of block
Therefore V= (1000- 200)/ 9 = 8.889

Kinetic energy = 1/2 x mass x veloc^2
Therefore:
Energy before impact = 1/2 x 0.01x 100^2
Energy after impact = 1/2 x 0.01x 20^2 +1/2 x 0.09x 8.889^2
Energy difference (loss) = 44.4 J

Note that only momentum is concerved. Energy can be lost in friction etc.

Answer 3

a. Work = Net Force * Distance.

Horizontal component of Force of worker=80cos60=40N
The Net Force= 40-24=16N
The Distance=6
Work =16*6=96J

b. the answer is a.

The work problem is self-explanatory. The rocket problem is about conservation of momentum. The mass of the fuel multiplied by its velocity is conserved, i.e. it comes as equal to the mass of the rocket multiplied by its velocity.

The bullet problem:
It's initial kinetic energy =1/2mv^2=1/2*10/1000*100^2=50 J

It's final kinetic energy=1/2*10/1000*20^2=2 J

Energy lost as it went through the block=50-2=48J