Question

A 2200 kg truck has put its front bumper against the rear bumper of a 2400 kg SUV to give it a push. With the engine at full power and good tires on good pavement, the maximum forward force on the truck is 18,000 N.
1. What is the maximum possible acceleration the truck can give the SUV?
2. At this acceleration, what is the force of the SUV’s bumper on the truck’s bumper?

Answer 1

Answer

given,

mass of truck (m_t)= 2200 Kg

mass of SUV (m_s)= 2400 Kg

Maximum forward force on the truck = 18,000 N.

a) $F = (m_t + m_s) a _{max}$

 both the body will be considered as a system

  $a _{max}= \dfrac{F}{(m_t + m_s)}$

  $a _{max}= \dfrac{18000}{(2200+2400)}$

  $a _{max}= \dfrac{18000}{4600}$

  $a _{max}= 3.91\ m/s^2$

b) Force of the SUV’s bumper on the truck’s bumper

  F = m_s a

  F = 2400 x 3.91

  F = 9384 N