Q.Read the following passage and answer the questions that follow. Meera, a 60 kg sprinter, settles into the starting blocks wearing spiked shoes. At the gun she drives her feet backward and downward against the block pads and explodes down the synthetic track. Assume the coefficient of static friction between her spikes and the track is 0.8 and take g = 9.8 m/s².

(i)

Newton's Third Law of Motion explains why Meera moves forward when she pushes backward on the blocks. The law states that for every action there is an equal and opposite reaction.

The action force is the force Meera exerts backward and downward on the starting blocks (and the track). The reaction force is the equal and opposite force the blocks (and track) exert forward and upward on Meera, propelling her down the track.

(ii)

The maximum static friction force is given by

fₛ^(max) = μₛ N.

When Meera is standing on the track, the normal reaction N equals her weight:

N = mg = 60 × 9.8 = 588 N.

Substituting into the friction formula,

fₛ^(max) = 0.8 × 588 = 470.4 N.

The maximum horizontal friction force available to her is 470.4 N.

(iii)

Her spikes help by increasing the effective coefficient of static friction between her shoes and the track. The sharp points dig into the synthetic surface, creating a mechanical interlock that prevents slipping and allows her to generate a larger backward push (and thus a larger forward reaction force) without her feet sliding out.

If the track were covered with a thin film of oil, the coefficient of friction would drop dramatically. She would be unable to generate sufficient friction force; her feet would slip backward on the blocks and track, and she would lose the explosive forward acceleration needed for a fast start.

(iv)

As she rises onto the balls of her feet at push-off, a second-class lever operates at the ankle.

• Fulcrum: the ball of the foot (metatarsophalangeal joints), where the foot contacts the ground. …

(i) Law of motion, action force and reaction force

Newton's Third Law of Motion governs Meera's forward acceleration. This law states that for every action there is an equal and opposite reaction — forces always occur in pairs acting on different bodies.

When Meera pushes backward and downward against the starting blocks, that push is the action force (her foot exerts a force on the blocks). The blocks, being anchored to the track, push forward and upward on her feet with an equal magnitude — this is the reaction force (the blocks exert a force on her). It is the reaction force that accelerates her body forward down the track.

The action and reaction are equal in size, opposite in direction, and crucially act on different objects. The action acts on the blocks; the reaction acts on Meera. Because Meera has much less mass than the Earth-track-block system, she accelerates forward while the blocks remain stationary.

(ii) Maximum horizontal friction force

The maximum static friction force determines how hard Meera can push without her spikes slipping.

where μₛ is the coefficient of static friction and N is the normal reaction force.

Working:

1. Normal reaction when standing upright:

   N = mg = 60 × 9.8 = 588 N

2. Maximum static friction:

   fₛ^(max) = μₛ N = 0.8 × 588 = 470.4 N

So the track can supply up to 470.4 N of horizontal friction force before her spikes slip. This is the upper limit on the horizontal propulsive force she can generate during the start.

(iii) Role of spikes and effect of oil

Spikes increase the coefficient of static friction by penetrating the synthetic surface and creating mechanical interlocking between the shoe and the track. The small contact area concentrates her weight, allowing the metal pins to grip the track material rather than simply sliding over it. This raises μₛ well above what a smooth rubber sole would achieve, permitting a much larger backward push without slipping.

If the track were covered with a thin film of oil, the coefficient of friction would drop dramatically — oil acts as a lubricant, reducing μₛ to perhaps 0.1 or less. Her spikes would slip during the explosive push-off, converting what should be forward acceleration into wasted sliding motion. She would lose traction, her start would be slow and unstable, and she would risk falling.

(iv) Lever class at the ankle during push-off

As Meera rises onto the balls of her feet at push-off, the ankle operates as a second-class lever.

In a second-class lever the load lies between the fulcrum and the effort: …