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.