Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

(a) The force of 4500 N will accelerate the system of the car and truck. We can find the acceleration of the system. $a = \frac{F}{m_c+m_t}$ $a = \frac{4500~N}{1000~kg+2000~kg}$ $a = 1.5~m/s^2$ The force $F_{ct}$ that the car exerts on the truck provides the force to accelerate the truck with an acceleration of $1.5~m/s^2$. We can find $F_{ct}$ $F_{ct} = m_t~a$ $F_{ct} = (2000~kg)(1.5~m/s^2)$ $F_{ct} = 3000~N$ The magnitude of the force of the car on the truck is 3000 N (b) According to Newton's third law, the force of the truck on the car is equal and opposite to the force of the car on the truck. Therefore, the magnitude of the force of the truck on the car is 3000 N