Question

A 12.45 kg block is slid from a height of 12.55 m down a friction-less incline. It comes to a flat surface and crosses a 6.67 m strip of sandpaper with a coefficient of friction of 0.34. After passing over the sandpaper, the block is traveling on a friction-less surface again. How fast is the block traveling? 

Answer 1

Since the incline is friction-less, we can use the energy conservation law for the first part of the motion. The potential energy of the block at the beginning of the motion is equal to its kinetic energy before it enters the strip of sandpaper:

$$E_p = E_{k0}$$

On the other hand, the potential energy is given as follows:

$$E_p = mgh$$

where $m = 12.45kg$ is the mass of the block, $g = 9.81m/s^2$ is the gravitational acceleration, and $h = 12.55m$ is the initial height. Thus, obtain:

$$E_{k0} = E_p = mgh$$

During the second part of the motion, the strip of sandpaper does some work to the block. According to the work-energy theorem, the change in block's kinetic energy is equal to the net work $W$ done by the strip of sandpaper:

$$E_k-E_{k0} = W$$

where $E_{k0} = mgh$ is the kinetic energy of the block right before it enters the strip, and $E_k$ is its kinetic energy right after it leaves it.

On the other hand, by definition, the work is (negative, since sandpaper tries to stop the block):

$$W = -Fs$$

where $s = 6.67m$ is the length of the strip, and $F$ is the friction force:

$$F = \mu mg$$

where $\mu = 0.34$ is the coefficient of friction.

Combining it all together, obtain:

$$E_k = -\mu mgs + mgh = mg( h-\mu s)$$

The kinetic enegry at the left hand side can be writen as:

$$E_k =\dfrac{mv^2}{2}$$

where $v$ is the velocity of the block after it leaves the strip. Expressing the velocity, obtain:

$$\dfrac{mv^2}{2} = mg( h-\mu s)\\ v = \sqrt{2g(h-\mu s)}$$

Substituting numerical values, obtain:

$$v = \sqrt{2\cdot 9.81\cdot ( 12.55-0.34\cdot 6.67 )} \approx 14.2m/s$$

Answer. 14.2 m/s.