An 8 kg block, P, is being pulled by constant force F up a rough inclined plane at an angle of 30° to the horizontal, at CONSTANT SPEED - NSC Physical Sciences - Question 2 - 2017 - Paper 1

In the free-body diagram for block P, label the following forces:

• Weight (W) acting downward due to gravity.
• Normal force (N) acting perpendicular to the inclined plane.
• Friction force (f_k) acting opposite to the direction of motion.
• Applied force (F) acting parallel to the inclined plane.

To find the magnitude of force F, we can use the equation for net force:
$F_{net} = ma$
At constant speed, the net force is zero, so:
$F - f_k - F_g imes ext{sin}(30°) = 0$
Where:

• $f_k = 20.37$ N (the kinetic frictional force).
• The weight of the block is $W = mg = (8 ext{ kg}) imes (9.81 ext{ m/s}^2) = 78.48 ext{ N}$.
  Then,
  $F - 20.37 - (78.48 imes ext{sin}(30°)) = 0$
  $F = 20.37 + 39.24 = 59.61 ext{ N}$.

With force F removed, the block accelerates down the plane. The net force acting on the block is:
$F_{net} = mg imes ext{sin}(30°) - f_k$
This leads to:
$F_{net} = (8 ext{ kg} imes 9.81 ext{ m/s}^2 imes ext{sin}(30°)) - 20.37$
Calculating gives:
$F_{net} = 39.24 ext{ N} - 20.37 ext{ N} = 18.87 ext{ N}$
Now, using Newton's second law, the acceleration (a) can be calculated as:
$$a = rac{F_{net}}{m} = rac{18.87}{8} ext{ m/s}^2 = 2.36 ext{ m/s}^2.$