Question

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Problem 5.40 A 2200 kg truck has put its front bumper against the rear bumper of a 2500 kg SUV to give a push. With the engine at full power and good tires on pavement the maximum formal- force on the truck is 18,000N. Pert A.What is the maximum possible acceleration the truck can give the SUV? Express your answer to two significant and include the appropriate units. Part B At this acceleration, what is the force of the SUV, bumper on the truck bumper,? Express your answer to two significant and include the appropriate units.

Answer 1

Concepts and reason

The concept of Newton’s second law and Newton’s third laws are to be used to solve the given problem.

First, find the expression of maximum acceleration the truck give to the SUV using Newton’s second law and then use the value of acceleration to find the force of the SUV’s bumper on the truck’s bumper.

Fundamentals

Newton’s second law states that the force is equals to the rate of change in momentum. It also states that the net force acting on a body must be zero when the system is in equilibrium.

Newton’s third law states that for every action there is equal and opposite reaction.

The expression of the acceleration of a body is given as follows:

$a = \frac{F}{m}$

Here, F is the force applied and m is the mass of body.

(A)

From Newton’s second law, the net force on the system of truck and the SUV is,

${F_{\max }} = \left( {{m_{truck}} + {m_{SUV}}} \right){a_{\max }}$

Here, ${m_{SUV}}$ is the mass of SUV, ${m_{truck}}$ is the mass of truck, and ${a_{\max }}$ is the maximum acceleration of the total system of truck and the SUV.

Rearrange the above equation for ${a_{\max }}$ .

${a_{\max }} = \frac{{{F_{\max }}}}{{{m_{truck}} + {m_{SUV}}}}$

Substitute 18000 N for ${F_{\max }}$ , 2200 kg for ${m_{truck}}$ and 2500 kg for ${m_{SUV}}$ in equation ${a_{\max }} = \frac{{{F_{\max }}}}{{{m_{truck}} + {m_{SUV}}}}$ .

$\begin{array}{c}\\{a_{\max }} = \frac{{18000{\rm{ N}}}}{{2200{\rm{ kg}} + 2500{\rm{ kg}}}}\\\\ = 3.83{\rm{ m/}}{{\rm{s}}^2}\\\end{array}$

(B)

The force exerted by the SUV’s bumper on the truck’s bumper is calculated as follows:

$F = {m_{SUV}}{a_{\max }}$

Substitute 2500 kg for ${m_{SUV}}$ and $3.83{\rm{ m/}}{{\rm{s}}^2}$ for ${a_{\max }}$ in the above equation.

$\begin{array}{c}\\F = \left( {2500{\rm{ kg}}} \right)\left( {3.83{\rm{ m/}}{{\rm{s}}^2}} \right)\\\\ = 9575{\rm{ N}}\\\end{array}$

Ans: Part A

The maximum possible acceleration of the truck is $3.83{\rm{ m/}}{{\rm{s}}^2}$ .

Part B

The force of the SUV’s bumper on the truck’s bumper is 9575 N.