First, I'm sorry for asking a lot of questions, however I don't know how to solve any of these problems. Can you please help me by telling me how I should solve it and what formulas should be used? I don't want the actual answers but how to solve these problems. Please and thank you.

Question

1. A 15 kg block is moving at 5 m/s on a horizontal track. The track then dips down, before finally rising up. The height is 3 m. The surface is friction-less for all parts of this problem.
a. How fast is the block moving at the lowest part of the track?
b. How high is the block when it is moving up the far side of the track at 2 m/s?
c. What maximum height will the block reach on the far side of the track?

2. A piece of string of length 0.8 m is attached at one end to support rod and at the other to a ball with a mass of 2 kg. The ball is held so that the string makes an angle of 50 degrees tot he vertical, and then released from rest. After being released the ball oscillates back and forth.
a. What is the initial height of the ball above its lowest position?
b. After it is released, where does the ball have its maximum speed? What is the maximum speed of the ball?
c. What is the height of the ball above its lowest position when it is moving at one half of its maximum speed?

3. A block of mass 25 kg is sliding at 10 m/s on an initially friction-less, horizontal surface. There is then a 2 m patch of roughness where friction is present (u= 0.4), beyond which the surface is once again friction-less.
a. What is the magnitude of the friction force between the block and the rough patch of the surface?
b. How much work is done by the friction force as the block moves across the rough patch of surface?
c. How fast is the block moving on the far side of the rough patch?
d. What would the length of the rough patch of surface have to be to bring the block to a complete stop?

Answer 1

3. M * g = 25 * 9.8 = 245 N. = Wt. of block = Normal force(Fn).

a. 0.4 * 245 = 98 N. = Force of kinetic friction.

b. W = 98 * 2 = Joules.

c. Force applied - Force of friction = M * a.
0 - 98 = 25a,
a = -3.92 m/s^2.
V^2 = Vo^2 + 2a * L = 0.
10^2 + 2(-3.92)L = 0,
100 - 15.7L = 0,
L = ?