Question

A 630 kg car pulling a 515 kg trailer accelerates forward at a rate of 2.18 m/s². Assume frictional forces on the trailer are negligible. Ignore air drag. (a) Calculate the net force (in N ) on the car. magnitude ∣N direction (b) Calculate the net force (in N ) on the trailer. magnitude direction (c) What is the force (in N ) exerted by the trailer on the car? magnitude ] N direction the direction in degrees measured from the left of the −y-direction.) magnitude N direction - (measured from the left of the −y-direction)

Answer 1

The net force on the 630 kg car is 1373.4 N forward, on the 515 kg trailer it is 1122.7 N forward, and the force exerted by the trailer on the car is 1122.7 N backward.

Step-by-step explanation:

To answer the student's question, we'll start by using Newton's second law of motion, which states that the force on an object is equal to its mass multiplied by its acceleration (F=ma).

For part (a), the net force on the car can be calculated as the mass of the car multiplied by the acceleration. So, the net force on the 630 kg car accelerating at 2.18 m/s2 is 1373.4 N in the direction of acceleration (forward).

For part (b), the net force on the trailer is calculated in the same way. Assuming there are no frictional forces acting on it, the force will be equal to the mass of the trailer multiplied by the acceleration. The net force on the 515 kg trailer is 1122.7 N, also in the direction of acceleration (forward).

For part (c), the force exerted by the trailer on the car (according to Newton's third law of motion, which states that every action has an equal and opposite reaction) will be equal in magnitude and opposite in direction to the force the car exerts on the trailer, which is 1122.7 N backward, or in the opposite direction to the car's acceleration.