Twee blokke met massas van 220 kg en 75 kg lê stil op 'n RUWE horisontale oppervlak - NSC Technical Sciences - Question 2 - 2023 - Paper 1

Newton's Second Law states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration:

$F_{net} = ma$. In this case,

$F_{net} = F_{applied} - F_{friction} - T_x$

where $F_{applied} = 165 ext{ N}$, $F_{friction} = 35 ext{ N}$, and $T_x$ is the horizontal component of tension. Since the block is moving to the left, we need to take into account the direction of each force.

Using the net force equation from Newton's Second Law, we have:

$F_{net} = 165 ext{ N} - 35 ext{ N} - T_x$. Assuming the acceleration (a) is constant and from the problem, we can figure out the effective mass as follows:

a = rac{F_{net}}{m} For our setup:

$F_{net} = 220 ext{ kg} imes 0.4 ext{ m}$ = mass x acceleration. Solving this yields T = 48.498 ext{ N} or approximately 48.28 ext{ N}.

To find the coefficient of kinetic friction ($μ_k$), we use:

$f_k = μ_k imes N$, where:

• $f_k$ = 12 N (given)
• $N$ = weight of the 75 kg block = $m imes g = 75 ext{ kg} imes 9.81 ext{ m/s}^2 = 735.75 ext{ N}$.

Rearranging the formula gives:

The tension in the rope affects the motion of the blocks. As more force is applied to move the 220 kg block, the interaction between frictional forces and the applied forces will dictate the overall movement. If the tension increases, it would generally suggest an increase in acceleration unless countered by resistance. Therefore, the understanding of tension as it impacts the blocks confirms how net forces change in response to applied external forces.