Question

VERY URGENT PLEASE HELP!!!

A block of mass 493 g is pushed against the
spring (located on the left-hand side of the
track) and compresses the spring a distance
4.9 cm from its equilibrium position (as shown
in the figure below). The block starts from
rest, is accelerated by the compressed spring,
and slides across a frictionless horizontal track
(as shown in the figure below). It leaves the
track horizontally, flies through the air, and
subsequently strikes the ground.
The acceleration of gravity is 9.81 m/s^2

a. what is the spring constant
b. what is the speed v of the block when it leaves the track?
c. what is the total speed of the block when it hits the ground?

imgur . com/a/XMjerGZ

Answer 1

It falls 2.2 meters from a Vi = initial speed down = 0
2.2 = (1/2) 9.81 t^2
t^2 = .449
t = 0.670 seconds in the air
Goes horizontal 3.99 meters in 0.670 seconds
so u = 3.99/.670 = 5.96 meters/ second
so Kinetic energy = Ke = (1/2)(.493)(5.96)^2
that was potential energy of spring = (1/2) k x^2
horizontal speed u = 5.96 still at ground
but vertical speed v = g t =9.81*.67
sqert(u^2+v^2)