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4. A 1,000 kg truck moving at 10 m/s runs into a concrete wall. It takes 0.5 seconds for the truck to conipietery
stop.
What is the magnitude of force exerted on the truck during the collision?
f. 5,000 N
h. 20,000 N
g. 10,000 N
j. 1,000 N

Respuesta :

znk

Answer:

[tex]\large \boxed{\text{h. 20 000 N}}[/tex] Ā 

Explanation:

Force is the change in momentum over time

F = Ī”p/Ī”t

1. Calculate the change in momentum

pā‚ = mvā‚ = 1000 kg Ɨ 10 m/s = 10 000 kgĀ·mĀ·sā»Ā¹

pā‚‚ = 0

Ī”p = pā‚‚ - pā‚= (0 - 10 000) kgĀ·mĀ·sā»Ā¹ = -10 000 kgĀ·mĀ·sā»Ā¹

2. Calculate the force

[tex]\begin{array}{rcl}F & = & \dfrac{\Delta p}{\Delta t}\\\\& = & \dfrac{-10 000 \text{ kg$\cdot$m$\cdot$ s}^{-1}}{\text{ 0.5 s}}\\\\& = & \textbf{-20 000 N}\\\end{array}\\\text{The negative sign shows that the force is exerted opposite to the direction of motion.}\\\text{The magnitude of the force is $\large \boxed{\textbf{20 000 N}}$}[/tex]

abidemiokin

The magnitude of the force exerted on the truck during the collision is 20,000N

The magnitude of the force exerted on the truck during the collision is expressed according to Newton's second law;

[tex]F = ma\\F=m(\frac{v-u}{t} )\\[/tex]

m is the mass of the object = 1000kg

v is the final velocity = 10m/s

u is the initial velocity = 0m/s

t is the time taken = 0.5secs

Substitute the given parameters into the formula:

[tex]F=1000(\frac{10-0}{0.5} )\\F=1000(\frac{10}{0.5})\\F=1000\times 20\\F=20,000N[/tex]

Hence the magnitude of the force exerted on the truck during the collision is 20,000N.

Learn more here:

https://brainly.com/question/1959798

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