A crate of mass 86 kg is accelerating down a surface inclined at an angle of 25° to the horizontal - NSC Physical Sciences - Question 3 - 2016 - Paper 1

To find the magnitude of the kinetic friction force (F_k), we use the formula:

$F_k = ext{μ}_k imes F_N$

Where:

• $ext{μ}_k = 0.22$ (coefficient of kinetic friction)
• $F_N = mg imes ext{cos}( heta)$, where $m = 86 ext{ kg}$, $g = 9.81 ext{ m/s}^2$, and $heta = 25°$.

Calculating $F_N$:

The work-energy theorem states that the work done on an object is equal to the change in its kinetic energy. In simpler terms, if a net work is done on the object, it results in a change in its speed or motion. Work done can be thought of as the transfer of energy to or from an object.

The free-body diagram should show the following forces acting on the crate:

• Weight (W) acting downward.
• Normal force (N) acting perpendicular to the inclined surface.
• Applied force (F) acting parallel to the inclined surface in the upward direction.
• Frictional force (F_f) acting downwards along the incline opposite to the direction of motion.

Label the forces accordingly for clarity.

Using the work-energy theorem, the work done (W) can be calculated as:

$W = F_{net} imes d$

Where:

• $F_{net} = m imes a$, with $m = 86 ext{ kg}$ and $a = 1.54 ext{ m/s}^2$.

Calculate $F_{net}$:

$F_{net} = 86 imes 1.54 = 132.64 ext{ N}$

Then calculate the work done over distance $d = 0.9 ext{ m}$:

$W = 132.64 imes 0.9 = 119.38 ext{ J}$

Thus, the work done is approximately 119.38 Joules.