Question

In the winter sport of curling, players give a 20 kgstone 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.

a. A curler pushes a stone to a speed of 3.0 m/s over a time of 1.4s . Ignoring the force of friction, how much force must the curler apply to the stone to bring it up to speed?
b. The sweepers in a curling competition adjust the trajectory of the stone by
c. Suppose that the stone is launched with a speed of 3 m/s and travels 40 m before coming to rest. What is the approximate magnitude of the friction force on the stone?
d. Suppose the stone's mass is increased to 40 kg, but it is launched at the same 3 m/s. Which one of the following is true?

Answer 1

(a) F = 42.86N

(c) Ff = –2.25N

(d) Ff = –4.50N

Step-by-step explanation:

(a) mass of stone m = 20kg

initial speed of stone u = 0m/s

final speed of stone v = 3.0m/s

Time interval = t = 1.4s

From Newtown's second law of motion

F = m(v – u)/t = 20(3.0 – 0)/ 1.4 = 42.86N

(c) v = 0 (comes to rest) u = 3m/s

v² = u² + 2aS

S = 40m

0² = 3³ + 2a×40

80a = –9

a = –9/80 = -0.1125

Once the stone is launched the only force acting on it is Ff the frictional force,

So

Ff = ma = 20×(–0.1125)

Ff = –2.25N

Negative sign shows the force is acting in a direction opposite the direction of motion.

(d) If the stone's mass is now 40kg

Ff = 40(–0.1125) = –4.45N

Answer 2

Questions B & D of the question are not complete and the complete question is;

B) The sweepers in a curling competition adjust the trajectory of the stone by A. Decreasing the coefficient of friction between the stone and the ice. B. Increasing the coefficient of friction between the stone and the ice. C. Changing friction from kinetic to static. D Changing friction from static to kinetic.

D) Suppose the stone's mass is increased to 40 kg, but it is launched at the same 3 m/s. Which one of the following is true? A. The stone would now travel a longer distance before coming to rest. B. The stone would now travel a shorter distance before coming to rest. C. The coefficient of friction would now be greater. D. The force of friction would now be greater.

Answer:

A) Force exerted on stone = 42.89 N

B) The correct option is A

C) Magnitude of the friction force = 2.25N

D) The correct option is D

Step-by-step explanation:

A) The first equation of motion is given as;

v = u + at

from the question,

v = 3 m/s

u = 0 m/s

t = 1.4 s

Acceleration (a) is unknown.

Thus, making a the subject,

a = (v-u)/t = (3 - 0)/1.4 = 2.1429 m/s²

Force is given as; F = ma

Thus, Force exerted on stone = 20 x 2.1429 = 42.89 N

B) The correct option is A because since friction opposes the relative motion of any object, thus when it is decreased, the object can travel some more distance.

C) Kinetic energy equation is given as;

Ek = (1/2)mv² = (1 /2)(20 x 3²) = 90 J

Now, work done = Force x Distance.

Thus, Frictional force on stone = Work done/Distance

Thus,

F = 90/40 = 2.25N

D) Formula for frictional force with coefficient of friction is given as;

F = μmg

Now, F is also known to be F = ma

Thus,

ma = μmg

So, a = μg

Where μ is coefficient of friction.

Inspecting a = μg, and the other equations of motion, none of the parameters changes with an increase in the mass except for frictional force. Hence, it means a, b, and c are wrong and answer choice D is correct. This is because when mass is increased, the overall frictional force applied is greater even though the distance remains the same.