A block of mass 5.8 kg is sitting on a frictionless ramp with a spring at the bo

Question

A block of mass 5.8 kg is sitting on a frictionless ramp with a spring at the bottom, as shown, that has a spring constant of 545 N/m. The angle of ramp with respect to the horizontal is 25 degree. The block, starting at rest, slides down the ramp a distance 34 cm before hitting the spring. How far, in centimeters, does the spring compress before the block stops? After the block comes to rest, the spring pushes the block back up the ramp. How fast, in meters per second, is the block moving right after it comes off the spring? What is the difference, in Joules, in gravitational potential energy between the original position of the block at the top of the ramp and the position of the block when the spring is fully compressed?

Explanation / Answer

m = 5.8 Kg
k = 545 N/m
= 25

(a)
Let the spring compression = x m
h = (0.34 + x)* sin(25)
Block will stop when all the gravitational potential Energy will be converted to Spring potential Energy.
Using Energy Conservation,

Initial Potential Energy = Final Spring Potential Energy
m*g*h = 1/2*kx^2
5.8*9.8*  (0.34 + x)* sin(25) = 1/2*545 * x^2
x = 0.223 m
Spring compression, x = 22.3 cm

(b)
As there is no friction, All the spring potential Energy will be converted to Kinetic Energy + Potential Energy gained.
1/2 * kx^2 = 1/2*mv^2 + m*g*x*sin(25)
545*0.223^2 = 5.8*v^2 + 5.8*9.8*0.223*sin(25)
v = 1.94 m/s
Speed of the block right after it comes off the spring,V = 1.94 m/s

(c)
We will take Gravitational Potential Energy at the bottom when spring is fully compressed = 0
Gravitational Potential Energy at the top of the ramp, = 5.8*9.8*(0.34 + 0.223)*sin(25) J
Gravitational Potential Energy at the top of the ramp, = 13.5 J
So,
Difference, = 13.5 J

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