Question

A box is given an initial speed of u on a rough horizontal surface of coefficient of kinetic friction , . Find the
distance the box travels before coming to rest.
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Answer 1

Question :

A box is given an initial velocity of u on a rough horizontal surface of coefficient of kinetic friction μₖ Find the distance the box travels before coming to rest.

Given :

Initial velocity = u

Coefficient of kinetic friction = μₖ

To Find :

Distance covered by block before coming to rest.

Solution :

❖ We know that frictional force opposes the relative motion of body with respect to ground in other words it produces retardation or better to say negative acceleration in the body.

» Here frictional force acts in the opposite direction of motion$.$

➙ Force = mass × acceleration

➙ μₖ × N = m × a

• N denotes normal reaction (= mg)

➙ μₖ × mg = m × a

➙ a = - μₖ g

Negative sign shows that velocity of block decreases with time.

Applying third equation of kinematics;

➠ v² - u² = 2as

➠ 0² - u² = 2 (-μₖ g) s

➙ u² = 2 μₖ g s

➙ s = u² /2 μₖg

∴ (A) is the correct answer!

Answer 2

Question :

A box is given an initial velocity of u on a rough horizontal surface of coefficient of kinetic friction μₖ Find the distance the box travels before coming to rest.

Given :

Initial velocity = u

Coefficient of kinetic friction = μₖ

To Find :

Distance covered by block before coming to rest.

Solution :

❖ We know that frictional force opposes the relative motion of body with respect to ground in other words it produces retardation or better to say negative acceleration in the body.

» Here frictional force acts in the opposite direction of motion$.$

➙ Force = mass × acceleration

➙ μₖ × N = m × a

N denotes normal reaction (= mg)

➙ μₖ × mg = m × a

➙ a = - μₖ g

Negative sign shows that velocity of block decreases with time.

Applying third equation of kinematics;

➠ v² - u² = 2as

➠ 0² - u² = 2 (-μₖ g) s

➙ u² = 2 μₖ g s

➙ s = u² /2 μₖg

∴ (A) is the correct answer!