Question

A puck of mass 30.0kg slides across rough ice, experiencing a frictional force of 0.20 N. If it was moving at 10.0 km/h when it hit the ice patch, how long did it take to stop?

Answer 1

Answer on Question #49168, Physics, Mechanics | Kinematics | Dynamics

A puck of mass $30.0\,\mathrm{kg}$ slides across rough ice, experiencing a frictional force of $0.20\,\mathrm{N}$. If it was moving at $10.0\,\mathrm{km/h}$ when it hit the ice patch, how long did it take to stop?

Solution:

The magnitude of force is equated to the product of the mass times the acceleration.

$$F = -ma$$

Thus, the acceleration is

$$a = -\frac{F}{m} = \frac{-0.20}{30.0} = -0.0067\, \mathrm{m/s^2}$$

The kinematics equation is

$$a = \frac{v_f - v_i}{t}$$ $$v_f = 0$$ $$v_i = 10.0 \times \frac{1000}{3600} = 2.78\, \mathrm{m/s}$$

Thus, time is

$$t = \frac{-v_i}{a} = \frac{-2.78}{-0.0067} = 414.9\, \mathrm{s}$$

Answer: $t = 414.9\,\mathrm{s}$

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