Grade 11: Physics worksheet: Vectors in two dimensions Knowledge Area: Mechanics Draw a labelled force diagram and free body diagram for the all of the questions Question 1 A crate is pushed with a force of 40 N to the left. The crate is at rest. Calculate the frictional force acting on this object. Question 2 A crate of mass 50 kg is pushed at a force of 800 N to the right. The crate is moving. The coefficient of kinetic friction is 0,74 . Calculate the frictional force acting on the crate. Question 3 A box of 6 kg is on an inclined surface that is \( 25^{\circ} \) with the horizontal. The crate is about to move. There is a frictional force of 25 N present between the box and surface. Calculate the coefficient of static friction \( \left(\mu_{\mathrm{s}}\right) \) for these two surfaces. Question 4 A suitcase of 23 kg is pulled with a force of 54 N at an angle of \( 43^{\circ} \) with the vertical. The suitcase is pulled at a constant velocity. Calculate the frictional force. Question 5 A car \( (1200 \mathrm{~kg}) \) is being pulled up a slope. The slope has an angle of \( 38^{\circ} \) with the horizontal. The frictional force present is 1000 N . The car moves at a constant velocity. Calculate the applied force. Question 6 A toy of 200 g is pushed down a slope \( 23^{\circ} \) with the horizontal. The applied force is 15 N . The toy accelerates at \( 10.5 \mathrm{~m} . \mathrm{s}^{-2} \). Calculate the frictional force. Question 7 There is a pushing force at an angled of \( 42^{\circ} \) with the vertical acting on an object with a weight of 39 N . The applied force is 47 N . Calculate the normal force acting on the object.

To solve the problems presented, we will follow a systematic approach for each question, including drawing a labeled force diagram and free body diagram, and performing the necessary calculations step by step. ### Question 1 **Given:** - Force applied \( F = 40 \, \text{N} \) (to the left) - The crate is at rest. **Free Body Diagram:** - The force of friction \( f \) acts to the right (opposite to the applied force). - The crate is in equilibrium, so the net force is zero. **Calculation:** Since the crate is at rest, the frictional force must equal the applied force: \[ f = F = 40 \, \text{N} \] ### Question 2 **Given:** - Mass of crate \( m = 50 \, \text{kg} \) - Force applied \( F = 800 \, \text{N} \) (to the right) - Coefficient of kinetic friction \( \mu_k = 0.74 \) **Free Body Diagram:** - The applied force \( F \) acts to the right. - The frictional force \( f_k \) acts to the left. **Calculation:** 1. Calculate the normal force \( N \): \[ N = m \cdot g = 50 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 = 490.5 \, \text{N} \] 2. Calculate the frictional force: \[ f_k = \mu_k \cdot N = 0.74 \cdot 490.5 \, \text{N} = 363.87 \, \text{N} \] ### Question 3 **Given:** - Mass of box \( m = 6 \, \text{kg} \) - Angle of incline \( \theta = 25^{\circ} \) - Frictional force \( f_s = 25 \, \text{N} \) **Free Body Diagram:** - The weight \( W = m \cdot g \) acts downward. - The normal force \( N \) acts perpendicular to the surface. - The frictional force \( f_s \) acts up the incline. **Calculation:** 1. Calculate the weight: \[ W = m \cdot g = 6 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 = 58.86 \, \text{N} \] 2. Calculate the normal force: \[ N = W \cdot \cos(\theta) = 58.86 \cdot \cos(25^{\circ}) = 52.78 \, \text{N} \] 3. Calculate the coefficient of static friction: \[ \mu_s = \frac{f_s}{N} = \frac{25}{52.78} \approx 0.474 \] ### Question 4 **Given:** - Mass of suitcase \( m = 23 \, \text{kg} \) - Force applied \( F = 54 \, \text{N} \) at \( 43^{\circ} \) with the vertical. **Free Body Diagram:** - The applied force has a vertical and horizontal component. - The frictional force \( f \) acts opposite to the direction of motion. **Calculation:** 1. Calculate the weight: \[ W = m \cdot g = 23 \cdot 9.81 = 225.63 \, \text{N} \] 2. Calculate the vertical component of the applied force: \[ F_y = F \cdot \cos(43^{\circ}) = 54 \cdot \cos(43^{\circ}) \approx 40.12 \, \text{N} \] 3. Calculate the normal force: \[ N = W - F_y = 225.63 - 40.12 = 185.51 \, \text{N} \] 4. Since the suitcase moves at constant velocity, the frictional force equals the horizontal component of the applied force: \[ F_x = F \cdot \sin(43^{\circ}) \approx 36.12 \, \text{N} \] Thus, the frictional force \( f = 36.12 \, \text{N} \). ### Question 5 **Given:** - Mass of car \( m = 1200 \, \text{kg} \) - Angle of slope \( \theta = 38^{\circ} \) - Frictional force \( f = 1000 \, \text{N} \) **Free Body Diagram:** - The weight \( W \) acts downward. - The applied force \( F_a \) acts up the slope. **Calculation:** 1. Calculate the weight: \[ W = m \cdot g = 1200 \cdot 9.81 = 11772 \, \text{N} \] 2. Calculate the component of weight down the slope: \[ W_{\text{down}} = W \cdot \sin(\theta) = 11772 \cdot \sin(38^{\circ}) \approx 7,174.56 \, \text{N} \] 3. Calculate the normal force: \[ N = W \cdot \cos(\theta) = 11772 \cdot \cos(38^{\circ}) \approx 9,267.56 \, \text{N} \] 4. The net force is zero (constant velocity): \[ F_a - W_{\text{down}} - f = 0 \implies F_a = W_{\text{down}} + f = 7174.56 + 1000 = 8174.56 \, \text{N} \] ### Question 6 **Given:** - Mass of toy \( m = 0.2 \, \text{kg} \) - Angle of slope \( \theta = 23^{\circ} \) - Applied force \( F = 15 \, \text{N} \) - Acceleration \( a = 10.5 \, \text{m/s}^2 \) **Free Body Diagram:** - The applied force acts down the slope. - The frictional force \( f \) acts up the slope. **Calculation:** 1. Calculate the weight: \[ W = m \cdot g = 0.2 \cdot 9.81 = 1.962 \, \text{N} \] 2. Calculate the component of weight down the slope: \[ W_{\text{down}} = W \cdot \sin(\theta) = 1.962 \