Question

A 2300 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.

Part A

What is the maximum possible acceleration the truck can give the SUV?

Express your answer to two significant figures and include the appropriate units.

a =

Part B

At this acceleration, what is the force of the SUV's bumper on the truck's bumper?

Express your answer to two significant figures and include the appropriate units.

Fsuv on truck =

Answer 1

Concepts and reason

The concept required to solve this problem is Newton’s second law of motion.

Initially, write an expression for the force according to the Newton’s second law of motion. Later, rearrange the expression for the acceleration. Finally, substitute the value of the acceleration obtained to find the new force.

Fundamentals

According to the Newton’s second law of motion, the net force is equal to the product of the mass and the acceleration of an object. The expression for the Newton’s second law of motion is as follows:

$F = ma$

Here, m is mass and a is the acceleration.

(a)

Rearrange the equation $F = ma$ for a.

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

Substitute 18,000 N for F and $\left( {2300{\rm{ kg + 2400 kg}}} \right)$ for m in the equation $a = \frac{F}{m}$ .

$\begin{array}{c}\\a = \frac{{18,000{\rm{ N}}}}{{\left( {2300{\rm{ kg + 2400 kg}}} \right)}}\\\\ = \frac{{18,000{\rm{ N}}}}{{\left( {4700{\rm{ kg}}} \right)}}\\\\ = 3.83{\rm{ m/}}{{\rm{s}}^2}\\\end{array}$

(b)

Substitute $3.83{\rm{ m/}}{{\rm{s}}^2}$ for a and 2400 kg for m in the equation $F = ma$ .

$\begin{array}{c}\\F = \left( {2400{\rm{ kg}}} \right)\left( {3.83{\rm{ m/}}{{\rm{s}}^2}} \right)\\\\ = 9120{\rm{ N}}\\\end{array}$

Ans: Part a

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

Part b

The net magnitude of the force of the SUV's bumper on the truck's bumper is 9120 N.