7. a 3.1 kg block is slid from a height of 9 m down a friction-less incline. it comes to a flat surface and…

Answer:

\boxed{12.5,\text{m/s}} (approximate value, detailed calculation below)

Explanation:

Step1: Calculate initial potential energy

The initial potential energy of the block is given by ( PE = mgh ), where ( m = 3.1,\text{kg} ), ( g = 9.8,\text{m/s}^2 ), and ( h = 9,\text{m} ).
( PE = 3.1 \times 9.8 \times 9 = 270.54,\text{J} ).

Step2: Calculate work done by friction

The work done by friction on the sandpaper strip is ( W_f = -\mu_k mgd ), where ( \mu_k = 0.22 ), ( d = 5,\text{m} ).
( W_f = -0.22 \times 3.1 \times 9.8 \times 5 = -33.658,\text{J} ).

Step3: Apply conservation of energy

The initial potential energy is converted into kinetic energy after the sandpaper, minus the work done by friction. Let ( v ) be the final velocity.
Initial energy (PE) + Work done by friction = Final kinetic energy (( KE ))
( 270.54 - 33.658 = \frac{1}{2}mv^2 )
( 236.882 = \frac{1}{2} \times 3.1 \times v^2 )

Step4: Solve for ( v )

First, multiply both sides by 2: ( 473.764 = 3.1v^2 )
Then, divide by 3.1: ( v^2 = \frac{473.764}{3.1} \approx 152.827 )
Take the square root: ( v = \sqrt{152.827} \approx 12.5,\text{m/s} ).