The only force opposing the block's sliding as it slows down is friction with magnitude <em>f </em>. By Newton's second law, the net force in this direction is
∑ <em>F</em> = -<em>f</em> = <em>ma</em> = (4.00 kg) <em>a</em>
Assuming constant acceleration <em>a </em>, the acceleration applied by friction is such that
(1.5 m/s)² - (11 m/s)² = 2<em>a</em> (4.00 m)
Solve for the acceleration :
<em>a</em> = ((1.5 m/s)² - (11 m/s)²) / (8.00 m) ≈ -14.8 m/s²
Then the frictional force exerted a magnitude of
-<em>f</em> = (4.00 kg) (-14.8 m/s²)
<em>f</em> ≈ 59.4 N
and was directed opposite the block's motion.