Question

a black of mass 10kg is placed on a rough horizontal surface having coefficient of friction u=0.5if a horizontal force of 100N is acting on it then acceleration of the block Will be​

Answer 1

Answer:

• The acceleration (a) of the block is 5 m/s²

Given:

1. Mass of the body (M) = 10 Kg
2. Coefficient of friction (μ) = 0.5
3. Applied Force (F) = 100 N

Explanation:

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Firstly Finding the Frictional force.

From the formula,

⇒ f = μ N

Here,

• f Denotes Frictional Force,
• μ Denotes coefficient of friction
• N Denotes Normal Reaction.

Substituting the values,

⇒ f = 0.5 × N

⇒ f = 0.5 × (M g)  ∵[N = m g]

⇒ f = 0.5 × (10 × 10)

⇒ f = 0.5 × 100

⇒ f = 50

⇒ f = 50 N

∴ We got the Frictional Force value.

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Applying Newton's Second law of motion.

⇒ F_{net} = M a

Here,

• F_{net} Denotes Net Force.
• M Denotes Mass.
• a Denotes Acceleration.

Substituting the values,

⇒ ( F - f ) = M a

⇒ ( 100 - 50 ) = 10 × a

⇒ 50 = 10 a

⇒ 10 a = 50

⇒ a = 50 / 10

⇒ a = 5

⇒ a = 5 m/s²

∴ The acceleration (a) of the block is 5 m/s².

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Answer 2

$\huge\underline\mathbb\red{Answer:-}$

the acceleration (a) of the block is 5m/s