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Home/Questions/A 2300 kg truck has put its front bumper against the rear bumper of a 2400...
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lyytutoriaExpert
Asked: April 13, 20222022-04-13T09:01:45+00:00 2022-04-13T09:01:45+00:00In: Physics

A 2300 kg truck has put its front bumper against the rear bumper of a 2400…

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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 =

♦ Relevant knowledge
According to Newton’s Second law of motion the force applied to an object is determined by an equation of mass plus the speed for the subject. Mathematically speaking, F=ma, where

  • F is the force that acts on the object
  • A is the object’s acceleration
  • M is the weight that the item weighs
accelerationforce
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    Olivia O'Kelly
    2022-04-13T09:01:49+00:00Added an answer on April 13, 2022 at 9:01 am
    Concepts and Reason

    This problem can be solved by Newton’s second law.

    Initially, you will write an expression to describe the force using Newton’s second law. Later, you will need to modify the expression to calculate the acceleration. To find the new force, you can substitute the acceleration value.

    Fundamentals

    The Newton’s second law is that the net force equals the product of the object’s mass and acceleration. This is the Newton’s second law.

    F = ma

    m refers to mass, and refers to acceleration.

    (a) Rearrange equation 

    a = F ⁄ m

    Substitute 18,000 N to F and (2300 + 2400 kg) for m in the equation a = F ⁄ m

    a = 18,000 N ⁄ (2300 kg + 2400 kg)

       = 18,000 N ⁄ 47oo kg

       = 3.83 m/s2

    (b)

    Substitute 3.83 m/s2 for a and 2400 kg for m in the equation F = ma

    F =(2400kg) (3.83 m/s2) = 9120N

    Ans:

    Part A

    Maximum acceleration that the truck can give an SUV is 3.83 m/s2

    Part B

    The force exerted by the bumper of an SUV on a truck’s bumper is 9120N.

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