Question

A block of mass =3.05 kg

slides along a horizontal table with speed 0=3.50 m/s.
At =0,
it hits a spring with spring constant =85.00 N/m,
and it also begins to experience a friction force.

The coefficient of friction is given by =0.150.

Two illustrations of a horizontal box spring system.

The top illustration shows a box traveling to the right at velocity v subscript 0 on a horizontal surface towards a relaxed spring. The spring rests on the surface and is attached to a wall on its right end.

The bottom illustration shows the box at rest against the left end of the spring. The spring has been compressed by an amount delta x.

How far Δ
has the spring compressed by the time the block first momentarily comes to rest?

Answer 1

0.612 m

Step-by-step explanation:

Initially, the block has kinetic energy, KE.

When the block stops, that energy is transferred into elastic energy EE and work done by friction W.

Energy is conserved:

KE = EE + W

½ mv² = ½ kx² + Fx

Friction force equals normal force times coefficient of friction:

½ mv² = ½ kx² + Nμ x

Sum of forces in the y direction:

∑F = ma

N − mg = 0

N = mg

Substituting:

½ mv² = ½ kx² + mgμ x

Substitute values:

½ (3.05 kg) (3.50 m/s)² = ½ (85.00 N/m) x² + (3.05 kg) (9.8 m/s²) (0.150) x

18.68 = 42.5 x² + 4.484 x

0 = 42.5 x² + 4.484 x − 18.68

Solve the quadratic equation:

x = [ -b ± √(b² − 4ac) ] / 2a

x = [ -4.484 ± √(4.484² − 4 • 42.5 • -18.68) ] / (2 • 42.5)

x = 0.612

The block moves 0.612 meters.