Vraag 3 \( \overline{\mathrm{r}} \) Blok rue on \( \hat{n} \) horioontake eppervilak. Dio nomaailkrag ios 20 N . Die Koeffigiont van atatiose wrywing tucoen die blok on dio opperviokiog 0,40 on dio kóóffisiont van dinamioese wryming ios 920. a) Nat is diegrootte van dio wrywingokeng wat op dio brok vifgeoofon word tomyl- die blok in rug is? b) Hat sal die grootte wan die wrytingetrag wees ac \( \hat{n} \) horifontalb krag Nan, grootte 5 N op die brok. 0) Hat is dio minmum krag rat benodig mord ón die blok to loat boncog. d) Hat ios dic minimum krog nat bonodig nord om die blok in bowroging the hou codra dit beginio?

Let's break down the problem step by step, extracting the known conditions and solving each part. ### Known Conditions: 1. Normal force (\( \hat{n} \)) acting on the block: \( 20 \, \text{N} \) 2. Coefficient of static friction (\( \mu_s \)): \( 0.40 \) 3. Coefficient of kinetic friction (\( \mu_k \)): \( 0.20 \) ### a) Calculate the magnitude of the frictional force acting on the block when the block is at rest. The maximum static frictional force (\( F_s \)) can be calculated using the formula: \[ F_s = \mu_s \cdot \hat{n} \] Substituting the known values: \[ F_s = 0.40 \cdot 20 \] Let's calculate this. Calculate the value by following steps: - step0: Calculate: \(0.4\times 20\) - step1: Multiply the numbers: \(8\) The maximum static frictional force acting on the block is \( 8 \, \text{N} \). ### b) Calculate the magnitude of the frictional force when a horizontal force of \( 5 \, \text{N} \) is applied to the block. Since the applied force (\( 5 \, \text{N} \)) is less than the maximum static frictional force (\( 8 \, \text{N} \)), the frictional force will equal the applied force to prevent motion. Therefore, the frictional force (\( F_f \)) is: \[ F_f = 5 \, \text{N} \] ### c) Calculate the minimum force required to move the block. To find the minimum force required to overcome static friction, we use the maximum static frictional force calculated earlier: \[ F_{\text{min}} = F_s = 8 \, \text{N} \] ### d) Calculate the minimum force required to keep the block in motion. Once the block is in motion, the kinetic frictional force (\( F_k \)) applies. This can be calculated using the formula: \[ F_k = \mu_k \cdot \hat{n} \] Substituting the known values: \[ F_k = 0.20 \cdot 20 \] Let's calculate this. Calculate the value by following steps: - step0: Calculate: \(0.2\times 20\) - step1: Multiply the numbers: \(4\) The kinetic frictional force acting on the block is \( 4 \, \text{N} \). Thus, the minimum force required to keep the block in motion is \( 4 \, \text{N} \). ### Summary of Answers: a) The magnitude of the frictional force when the block is at rest: \( 8 \, \text{N} \) b) The magnitude of the frictional force with a \( 5 \, \text{N} \) applied force: \( 5 \, \text{N} \) c) The minimum force required to move the block: \( 8 \, \text{N} \) d) The minimum force required to keep the block in motion: \( 4 \, \text{N} \)