• Net vertical force on the block:
∑ <em>F</em> = <em>n</em> - <em>w</em> = 0
(<em>n</em> = magnitude of normal force, <em>w</em> = weight)
<em>n</em> = <em>w</em> = <em>m g</em>
(<em>m</em> = mass, <em>g</em> = 9.8 m/s²)
<em>n</em> = (4 kg) (9.8 m/s²) = 39.2 N
• Net horizontal force:
∑ <em>F</em> = -<em>f</em> = <em>m a</em>
(<em>f</em> = mag. of friction, <em>a</em> = acceleration)
We have <em>f</em> = <em>µ</em> <em>n</em> = 0.5 (39.2 N) = 19.6 N, so
-19.6 N = (4 kg) <em>a</em>
<em>a</em> = -4.9 m/s²
With this acceleration, the block comes to a rest from an initial speed of 5 m/s, so that it travels a distance ∆<em>x</em> in this time such that
0² - (5 m/s)² = 2 (-4.9 m/s²) ∆<em>x</em>
∆<em>x</em> = (25 m²/s²) / (9.8 m/s²) ≈ 2.55 m ≈ 2.6 m
