In the winter sport of curling, two teams alternate sliding 20 kg stones on an icy surface in an attempt to end up with the stone closest to the center of a target painted on the ice. During one turn, a player releases a stone that travels 27.9 m before coming to rest. The friction force acting on the stone is 2.0 N. What was the speed of the stone when the player released it?

u = 2.36 m/s

Explanation:

Parameters given:

Mass of stone = 20kg

Distance moved before coming to rest = 27.9m

Frictional force = 2.0N

We can first find the acceleration of the stone by using the equation for frictional force:

F = -ma

Where m = mass of the object

a = acceleration

The negative sign denotes that the Frictional force acts opposite the motion of the stone.

=> 2.0 = -20*a

a = - 2.0/20

a = - 0.1m/s²

Using one of the equations of motion, we have that:

v² = u² + 2as

Where v = final velocity

u = initial velocity

s = distance moved.

In this case, v = 0m/s because the object comes to rest.

=> 0 = u² + (2 * - 0.1 * 27.9)

0 = u² - 5.58

u² = 5.58

u = √(5.58)

u = 2.36 m/s

The initial velocity of the stone is 2.36m/s

pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y