Question 55

1a. Reaction forces are perpendicular to the surface and frictional forces act to oppose motion, which in this case is down the plane.

1b. The block is not accelerating off or into the slope, but it is accelerating down the slope and has components of this acceleration which are vertical and horizontal.

1c. The force down the plane is $mgsin\alpha$ and the force up the plane is $\mu N$ but since the block is in equilibrium perpendicular to the slope, $N=mgcos\alpha$. Therefore we have that the force down the plane is $mgsin\alpha-\mu mgcos\alpha$. This is equal to Mass $\times$ Acceleration so we have:

$mgsin\alpha-\mu mgcos\alpha=ma$

$m$ cancels and the acceleration is constant so can be written as $\frac{v}{t}$ giving:

$v=gt(sin\alpha-\mu cos\alpha)$

1d. The presence of this drag force means that the above equation for acceleration is modified such that the resultant force is reduced by $kv$ to give $mgsin\alpha−\mu mgcos\alpha −kv=ma$. When the velocity is terminal, $a=0$. Substituting in and rearranging for $v$ yields:

$v=\frac{mg(sin\alpha−\mu cos\alpha)}{k}$

1a. Reaction forces are perpendicular to the surface and frictional forces act to oppose motion, which in this case is down the plane.

1b. The block is not accelerating off or into the slope, but it is accelerating down the slope and has components of this acceleration which are vertical and horizontal.

1c. The force down the plane is $mgsin\alpha$ and the force up the plane is $\mu N$ but since the block is in equilibrium perpendicular to the slope, $N=mgcos\alpha$. Therefore we have that the force down the plane is $mgsin\alpha-\mu mgcos\alpha$. This is equal to Mass $\times$ Acceleration so we have:

$mgsin\alpha-\mu mgcos\alpha=ma$

$m$ cancels and the acceleration is constant so can be written as $\frac{v}{t}$ giving:

$v=gt(sin\alpha-\mu cos\alpha)$

1d. The presence of this drag force means that the above equation for acceleration is modified such that the resultant force is reduced by $kv$ to give $mgsin\alpha−\mu mgcos\alpha −kv=ma$. When the velocity is terminal, $a=0$. Substituting in and rearranging for $v$ yields:

$v=\frac{mg(sin\alpha−\mu cos\alpha)}{k}$