Question

a fire man weighting 80kg slides down a vertical pole.if the risisting force of friction is constant value of 720N,his acceleration is m/s^2 is(g=10m/s^2)​

Answer 1

Given :-

▪ A fire man of mass, m = 80 kg slides down a vertical pole. The force of friction is constant value of -720 N (in opposite direction of motion)

▪ Acceleration due to gravity, g = 10 m/s²

To Find :-

▪ Acceleration of the fire man.

Solution :-

Since, Friction is always applied against the motion motion so it is negative here. Now, The force of weight of the fire man will be down.

⇒ Weight = Mass × Acceleration due to gravity

⇒ Weight = 80 × 10

⇒ Weight = 800 N

The resultant force will be downwards of magnitude 80 N. Here's why,

⇒ Resultant Force = Friction force + weight force

⇒ Resultant = -720 + 800

⇒ Resultant = 80 N

Now, The net downwards force is 80 N, we know

⇒ Force = Mass × Acceleration

⇒ 80 = 80 × a

⇒ a = 1 m/s²

Hence, The acceleration of the fire man will be 1 m/s² downwards.

Answer 2

Given:

• Mass= 80kg
• Resisting Force = 720N
• g= 10m/s2

━━━━━━━━━━━━━━━━━

Need to find:

• Acceleration =?

━━━━━━━━━━━━━━━━━

Solution:

The Force acting downwards = Mass × Acceleration due to gravity

⟹ Force = 80 × 10 N

⟹ Force = 800 N

━━━━━━━━━━

Now there is resisting Force acting in opposite direction

So net Force = 800N - 720 N

⟹ Net Force = 80N

━━━━━━━━━━

The Net Force= Mass × Acceleration of man

⟹ 80 = 80 × Acceleration

⟹ Acceleration = 80÷80

⟹$\red{\large{\bold{\boxed{Acceleration = 1m/ s^{2}}}}}$