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An 800-kg car is going 15m/sec along a level highway. How much force is required to stop the car in a distance of 60 meters?​

Sagot :

[tex]\tt{\huge{\blue{Explanation:}}}[/tex]

The force acting on an object is given by

[tex]\boxed{F = ma}[/tex]

where:

F = force

m = mass

a = acceleration

[tex]\tt{\huge{\red{Solution:}}}[/tex]

Solving for vf

[tex]v_{f}^{2} - v_{i}^{2} = 2aS[/tex]

[tex]\dfrac{v_{f}^{2} - v_{i}^{2}}{2S} = \dfrac{2aS}{2S}[/tex]

[tex]a = \dfrac{v_{f}^{2} - v_{i}^{2}}{2S}[/tex]

[tex]a = \dfrac{0^{2} - (\text{15 m/s})^{2}}{2(\text{60m})}[/tex]

a = -1.875 m/s²

Solving for F

F = ma

F = (800 kg)(-1.875 m/s²)

[tex]\boxed{F = -\text{1,500 N}}[/tex]

Therefore, the force required to stop the car is 1,500 N.

Note: The negative sign of the acceleration and the force indicates that the object is decelerating, meaning, the object is slowing down.

[tex]\\[/tex]

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