Question

A box of mass 10 kg pushed along rough floor with a velocity 1m\s and then let go The box moves 5 m on the floor before coming to rest. Then the force exerted on the object to bring it to rest is (A) -4N (B) -1N (C) -20 N (D) -8N​ brainliest for quickest answer fast!!!!!!!!!!!!!!!!!!!!!

Answer 1

Answer:

Mass of the box M = 10 kg

Initial Velocity u = 1 m/s

Final velocity v = 0

Distance travelled s = 5 m

v²-u²=2as

1 - 0 =2×a×5

a=1/10 m/s²

F=ma

=10×1/10

= 1 N

As the acceleration is negative ( deceleration of the object)

So -1N is the correct answer

Explanation:

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Answer 2

Given :

• Mass (m) = 10 kg
• Initial velocity (u) = 1 m/s
• Final velocity (v) = 0 m/s
• Distance travelled (s) = 5 m

To Find :

• Force exerted on object

Solution :

⇒v² - u² = 2as

⇒0² - 1² = 2*a*5

⇒10a = -1

⇒a = -1/10

⇒a = -0.1

$\therefore$ Acceleration of box is - 0.1 m/s²

___________________

⇒F = ma

⇒F = 10 * -0.1

⇒F = - 1 N

$\therefore$ Force exerted on object is - 1 N.