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A 100.2 g hockey puck sent sliding over ice is stopped in 15.1 m by the frictional force on it from the ice. If its initial speed is 6.54 m/s, what is the magnitude of the frictional force?

Answer :

Final answer:

The magnitude of the frictional force on the hockey puck from the ice is approximately 0.142 Newtons, calculated using the work-energy theorem and Newton's second law.

Explanation:

This question involves the concepts of Newton's second law, kinetic friction, and work-energy theorem. Given the mass of the hockey puck (m) is 100.2 g, the initial speed (u) is 6.54 m/s, and the distance (d) it stopped in is 15.1 m. The frictional force (f) can be calculated using the formula: f = (m*v²) / (2*d). Before that, we need to ensure that the mass is in kg, so it is 100.2 g = 0.1002 kg.

According to the work-energy theorem, the work done by the force is equal to the change in kinetic energy of the object. Since the puck is stopped, its final velocity (v) is 0, thus the change in kinetic energy is the initial kinetic energy which is (1/2*m*u²). Also, the work done by friction is the force of friction times the distance, so (f * d).

Setting them equal gives us: (1/2*m*u²) = (f * d). Solving this equation for the force of friction (f) yields to f = (0.1002 kg)*(6.54 m/s)² / (2*15.1 m) ≈ 0.142 N. Therefore, the magnitude of the frictional force on the puck from the ice is approximately 0.142 N.

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