In order get the block up to a speed of 3.58 m/s in 1.77 s, it must undergo an acceleration <em>a</em> of
<em>a</em> = (3.58 m/s) / (1.77 s) ≈ 2.02 m/s²
When the spring is getting pulled, Newton's second law tells us
• the net vertical force is
∑ <em>F</em> = <em>n</em> - <em>mg</em> = 0
where ∑ <em>F</em> is the net force, <em>n</em> is the magnitude of the normal force, and <em>mg</em> is the weight of the block - it follows that <em>n</em> = <em>mg</em> ; and
• the net horizontal force is
∑ <em>F</em> = <em>F</em> - <em>f</em> = <em>ma</em>
where <em>F</em> is the applied force, <em>f</em> is kinetic friction, <em>m</em> is the block's mass, and <em>a</em> is the acceleration found earlier. <em>F</em> stretches the spring by <em>x</em> = 0.250 m, so we have
<em>F</em> - <em>f</em> = <em>kx</em> - <em>µn</em> = <em>kx</em> - <em>µmg</em> = <em>ma</em>
where <em>k</em> is the spring constant and <em>µ</em> is the coefficient of kinetic friction.
When the block is being pulled at a constant speed, Newton's second law says
• the net vertical force is still
∑ <em>F</em> = <em>n</em> - <em>mg</em> = 0
so that <em>n</em> = <em>mg</em> again; and
• the net horizontal force is
∑ <em>F</em> = <em>F</em> - <em>f</em> = 0
This time, <em>F</em> stretches the spring by <em>y</em> = 0.0544 m, so we have
<em>F</em> - <em>f</em> = <em>ky</em> - <em>µmg</em> = 0
Solve the equations in boldface for <em>k</em> and <em>µ</em> :
<em>kx</em> - <em>µmg</em> = <em>ma</em>
<em>ky</em> - <em>µmg</em> = 0
==> <em>k</em> (<em>x</em> - <em>y</em>) = <em>ma</em>
==> <em>k</em> = <em>ma</em> / (<em>x</em> - <em>y</em>)
==> <em>k</em> = (17.6 kg) (2.02 m/s²) / (0.250 m - 0.0544 m) ≈ 182 N/m
Then
<em>ky</em> - <em>µmg</em> = 0
==> <em>µ</em> = <em>ky </em>/ (<em>mg</em>)
==> <em>µ</em> = (182 N/m) (0.0544 m) / ((17.6 kg) <em>g</em>) ≈ 0.0574
