A 1280-kg car pulls a 350-kg trailer. What force does the car exert on the trailer?

Calculation

A 1280-kg car pulls a 350-kg trailer. The car exerts a horizontal force of 3.6 × 10^3 N against the ground in order to accelerate. Assume an effective friction coefficient of 0.15 for the trailer.

Answer

Force exerted on the trailer: 1176 N

Explanation:

We are given that Mass of car (m1) = 1280 kg, Mass of trailer (m2) = 350 kg, Car exerts a horizontal force against the ground = 3.6 × 10^3 N, Coefficient of friction (μ) = 0.15.

We have to find the force exerted by the car on the trailer.

Using the formula F = (m1 + m2) × a + Ff, where Ff = μ × m2 × g and g = 9.8 m/s^2, we can calculate:

3.6 × 10^3 = (1280 + 350) × a + 0.15 × 350 × 9.8

1630a = 3600 - 514.5 = 3085.5

a = 3085.5 / 1630 = 1.89 m/s^2

Force exerted on the trailer:

F = m2 × a + μ × m2 × g

F = 350 × 1.89 + 0.15 × 350 × 9.8

F = 1176 N

Therefore, the car exerts a force of 1176 N on the trailer.

What is the force exerted by a 1280-kg car on a 350-kg trailer with a friction coefficient of 0.15 while accelerating with a force of 3.6 × 10^3 N?

The force exerted by the car on the trailer is 1176 N.