In the winter sport of curling, players give a 20 kg stone a push across a sheet of ice. the stone moves approximately 40 m before coming to rest. the final position of the stone, in principle, only depends on the initial speed at which it is launched and the force of friction between the ice and the stone, but team members can use brooms to sweep the ice in front of the stone to adjust its speed and trajectory a bit; they must do this without touching the stone. judicious sweeping can lengthen the travel of the stone by 3 m.

From Newton’s law of motion, the relation for velocity, acceleration and displacement is given by v^2 = vo^2 + 2as. Where v and vo are final and initial velocity respectively, and a, s are the acceleration and displacement respectively. From Newton’s second law, F = ma. Where m is the mass.

The magnitude of the frictional fore can be calculated only before estimating the ston’s acceleration. The initial speed of stone is mentioned in the problem as 3 m/s. Also, after 40m, the stone will stop. This means that its final speed is 0 m/s and the displacement is 40m. The acceleration of rock is calculated as:

(0 m/s) ^2 = (3 m/s) ^2 + 2a (40m).

9 m^2/s^2 = - 2a (40m)

a = - 9m^2/s^2/80m = 0.1 m/s^2.

This means that deceleration will occur and this will stop the stone. Substituting to Newton’s second law of motion,

-F = 20kg * - 0.1 m/s^2. Further solve,

F = 20kg * 0.1 m/s^2 = 2N. Hence, the magnitude frictional force is 2N.