Solve the physics problems in the image.

Steps to Solve

Here's a breakdown of how to solve each of the physics problems provided:

1. Force to Accelerate with Friction We need to calculate the force required to accelerate a 6kg object at 2m/s² on a surface with a kinetic friction coefficient of 0.4.

   First, calculate the frictional force: $F_{friction} = \mu \cdot N = \mu \cdot mg$, where $\mu$ is the coefficient of kinetic friction, $m$ is the mass, and $g$ is the acceleration due to gravity (approximately 9.8 m/s²). $F_{friction} = 0.4 \cdot 6 \cdot 9.8 = 23.52 N$

   Next, calculate the force required for the acceleration: $F_{acceleration} = ma = 6 \cdot 2 = 12 N$

   Finally, calculate the total force required: $F_{total} = F_{friction} + F_{acceleration} = 23.52 + 12 = 35.52 N$

2. Coefficient of Friction in Equilibrium A 4kg box is pulled by a 20N force and is in limiting equilibrium. This means the applied force equals the frictional force.

   $F_{friction} = \mu \cdot N = \mu \cdot mg = 20 N$ $\mu = \frac{20}{mg} = \frac{20}{4 \cdot 9.8} = \frac{20}{39.2} \approx 0.51$

3. Coefficient of Kinetic Friction with Angled Force A 3kg mass is pulled by a 60N force at 37º from the horizontal, accelerating at 5m/s².

   First, find the horizontal component of the applied force: $F_x = 60 \cdot \cos(37^\circ) \approx 60 \cdot 0.8 = 48 N$

   Next, find the net force causing acceleration: $F_{net} = ma = 3 \cdot 5 = 15 N$

   Then, find the friction force: $F_{friction} = F_x - F_{net} = 48 - 15 = 33 N$

   Now, find the normal force. It equals the weight minus the vertical component of the pulling force: $F_y = 60 \cdot \sin(37^{\circ}) \approx 60 \cdot 0.6 = 36 N$ $N = mg - F_y = 3 \cdot 9.8 - 36 = 29.4 - 36 = -6.6 N$. Since this cannot be negative, there must be something wrong. The acceleration of 5 m/s² must be lower for the $F_y$ to not become larger than the gravitational force. However, we will keep going with this value assuming that the prompt is correctly written.

   Lastly, calculate the coefficient of kinetic friction: $\mu = \frac{F_{friction}}{N} = \frac{33}{-6.6} = -5$ Since $\mu$ cannot be negative, it indicates that there is something wrong in the prompt and the numbers dont add up. However, we will keep going as if the absolute value were to be taken to the result.

4. Acceleration of Interconnected Masses We have three interconnected masses and $T_3 = 30N$. We want to calculate the acceleration of the system. Assume the masses are 2kg, 5kg, and 8kg.

Total mass $M = 2kg + 5kg + 8kg = 15kg$ The net force is $T_3 = 30N$

$F = Ma$ $a = \frac{F}{M} = \frac{30}{15} = 2 m/s^2$

5. Collision of Particles Particle A (2kg) moves at (3, 4) m/s and collides with particle B (1kg) moving at (6, 1) m/s. They stick together after the collision. We need to find their final velocity.

Use the conservation of momentum: $m_A v_A + m_B v_B = (m_A + m_B) v_{final}$ $(2kg)(3, 4) + (1kg)(6, 1) = (2kg + 1kg) v_{final}$ $(6, 8) + (6, 1) = 3 v_{final}$ $(12, 9) = 3 v_{final}$ $v_{final} = (\frac{12}{3}, \frac{9}{3}) = (4, 3)$ m/s

6. Collision of Pool Balls A pool ball (100g) with velocity (2, -4) m/s collides with a stationary pool ball of the same mass. After the collision, one ball has a velocity (1, -2) m/s. We need to find the velocity of the other ball.

Use the conservation of momentum: $m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f}$ Since $m_1 = m_2 = m$, we can divide through by $m$: $v_{1i} + v_{2i} = v_{1f} + v_{2f}$ $(2, -4) + (0, 0) = (1, -2) + v_{2f}$ $v_{2f} = (2, -4) - (1, -2) = (1, -2)$ m/s

7. Center of Mass We need to find the center of mass of the system from point A, given the masses and distances. The rod is massless.

$x_{cm} = \frac{\sum m_i x_i}{\sum m_i} = \frac{(3kg)(0m) + (3kg)(3m) + (3kg)(6m) + (2kg)(7m)}{3kg + 3kg + 3kg + 2kg}$ $x_{cm} = \frac{0 + 9 + 18 + 14}{11} = \frac{41}{11} \approx 3.73 m$

From point A, the center of mass is approximately 3.73 m.