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!!!!!!!!!!!!!!!!!!!!!
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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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Given :
To Find :
Solution :
⇒v² - u² = 2as
⇒0² - 1² = 2*a*5
⇒10a = -1
⇒a = -1/10
⇒a = -0.1
[tex]\therefore [/tex] Acceleration of box is - 0.1 m/s²
___________________
⇒F = ma
⇒F = 10 * -0.1
⇒F = - 1 N
[tex]\therefore [/tex] Force exerted on object is - 1 N.