Friendly Physics Advice requested on these Problems...?

How does one go about solving these...





1. At an amusement park, a swimmer uses a water slide to enter the main pool. If the swimmer starts at rest, slides without friction, and decends through a vertical height of 2.50 m, what is her speed at the bottom of the slide?


m/s





2. To enter the main pool at an amusement part, a swimmer uses a water slide which has a vertical height of 2.89 m. Find her speed at the bottom of the slide if she starts with an initial speed of 0.990 m/s.


m/s





3. A 2.0 kg block slides with a speed of 1.1 m/s on a frictionless, horizontal surface until it encounters a spring.


(a) If the block compresses the spring 5.0 cm before coming to rest, what is the force constant of the spring?


N/m


(b) What initial speed should the block have to compress the spring by 1.6 cm?


m/s





4. Catching a wave, a 65 kg surfer starts with a speed of 1.3 m/s, drops through a height of 2.00 m, and ends with a speed of 8.2 m/s. How much nonconservative work was done on the surfer?


kJ





TY.Friendly Physics Advice requested on these Problems...?
For the first two questions:





Somewhere in your text books it should state the acceleration of gravity... I believe its 10.9 metres per second squared (m/s^2). No, my mistake, its 9.8 metres per second squared, which means 9.8 metres per second, every second.





The formula can be found at the wikipedia link below. Use the classic formula.


Remember, vertical and horizontal calculations should be separate and all variables should be converted to metres (m), metres per second (m/s), metres per second squared (m/s^2) and seconds (s).





For questions 3 and 4 you will need the formula F=MA


ie: force equals mass multiplied by acceleration.


and W=Fd


ie work = force multiplied by distance.





Good luck.Friendly Physics Advice requested on these Problems...?
1. P.E.g initial = K.E.final


mgh = (1/2)mv^2


gh = (1/2)v^2


v = sqrt[2gh]


v = sqrt[2(9.8 m/s^2)(2.50 m)]


v = 7 m/s ANS





2. P.E.initial + KEinitial = P.Efinal + KEfinal


mghi + (1/2)mvi^2 = mghf + (1/2)mvf^2


ghi + (1/2)vi^2 = ghf + (1/2)vf^2


(9.8)(2.89) + (1/2)(0.990)^2 = 9.8(0) + (1/2)(vf^2)


vf = sqrt[{(9.8)(2.89) + (1/2)(0.990)^2}2]


vf = 5.31 m/sec ANS





3. a) KE. = P.E.elastic


(1/2)(2kg)(1.1 m/s)^2 = (1/2)k(0.05m)^2


(2)(1.1)^2 = k(0.05)^2


k = (2)(1.1)^2/(0.05)^2


k = 968 N/m ANS


b) KE = P.E.elastic


(1/2)(2)(vi)^2 = (1/2)(968)(0.016)^2


(2)(vi)^2 = (968)(0.016)^2


vi = sqrt[(968)(0.016)^2/2]


vi = 1.32 m/s