Ablock of mass 0.08 kg is pushed against a spring with spring constant k=31 n/m. the spring is compressed 0.15 meters from its natural length. the block is released and it slides along a surface with coefficient of friction 0.4. how far does the block travel from the point at which is is released?

1.11 meters

Explanation:

As the spring is compressed, elastic potential energy is built up in the spring. The total elastic potential energy can be found using the following formula

Ep = 1/2 x k x s²          

where k = 31 N/m  (spring constant)

s = 0.15 m  (compression)

Ep = 3.4875 J

When the block of mass is released, the elastic potential energy (Ep) is converted to kinetic energy (Ek). From this we can find the initial velocity of the mass of block after release

Ek = 1/2 x m x u²    

where Ek = Ep = 3.4875J

m = 0.08 kg  (mass of block)

u = unknown (initial velocity)

u = 2.9526 m/s

Now that we know the initial velocity we need to find the deceleration of the mass of block due to friction. We will first find the force of friction from the following formula

F = ∪ x m x g          

where F = unknown (frictional force)

∪ = 0.4   (coefficient of friction)

m = 0.08 kg   (mass of block)

g = 9.81 m/s² (acceleration due to gravity)

F = 0.31392 N

From this force we calculate the deceleration based on the following formula

F = m x a                  

where F = 0.31392   (frictional force)

m = 0.08 kg   (mass of block)

a = unknown  (acceleration)

a = -3.924 m/s²      -

*the negative sign is due to this value being deceleration

Now to find the total distance traveled we use the equation for motion

v² = u² + 2as            

where  v = 0 (final velocity)

u = 2.9526 m/s (initial velocity

a = -3.924 m/s² (deceleration due to friction)

s = unknown (distance traveled)

s = 1.11 meters

v 154.9= 0.9 length length s little over 154.9cm