Question

1. 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 crosses a 5 m strip of sandpaper with a coefficient of friction of 0.22. after passing over the sandpaper, the block is traveling on a friction-less surface again. how fast is the block traveling?

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 \( h = 9\,\text{m} \).
\( PE = 3.1 \times 9.8 \times 9 = 272.34\,\text{J} \).

Step2: Calculate work done by friction

The frictional force is \( F_f = \mu N \), and on a flat surface \( N = mg \), so \( F_f = \mu mg \). The work done by friction is \( W_f = -F_f d \), where \( d = 5\,\text{m} \) and \( \mu = 0.22 \).
\( F_f = 0.22 \times 3.1 \times 9.8 = 6.6836\,\text{N} \)
\( W_f = -6.6836 \times 5 = -33.418\,\text{J} \).

Step3: Apply work - energy theorem

The work - energy theorem states that the net work done on an object is equal to the change in its kinetic energy, \( W_{net}=\Delta KE \). The initial kinetic energy \( KE_i = 0 \) (starts from rest), and the final kinetic energy \( KE_f=\frac{1}{2}mv_f^2 \). The net work done is the initial potential energy plus the work done by friction (since the potential energy is converted to kinetic energy and work is done against friction).
\( PE + W_f=\frac{1}{2}mv_f^2 - 0 \)
\( 272.34- 33.418=\frac{1}{2}\times3.1\times v_f^2 \)
\( 238.922 = 1.55v_f^2 \)
\( v_f^2=\frac{238.922}{1.55}\approx154.143 \)
\( v_f=\sqrt{154.143}\approx12.9\,\text{m/s} \).

Answer:

\boxed{12.9\,\text{m/s}} (approximate value, detailed calculation below)