A 1.2-kg mass is projected down a rough circular track (radius = 2.0 m)as shown.The speed of the mass at point A is 3.2 m/s,and at point B,it is 6.0 m/s.What is the change in mechanical energy done on the system between A and B by the force of friction?
A)−8.9 J
B)−7.3 J
C)−8.1 J
D)−6.6 J
E)−24 J

Question

A 1.2-kg mass is projected down a rough circular track (radius = 2.0 m)as shown.The speed of the mass at point A is 3.2 m/s,and at point B,it is 6.0 m/s.What is the change in mechanical energy done on the system between A and B by the force of friction? 
A)−8.9 J 
B)−7.3 J 
C)−8.1 J 
D)−6.6 J 
E)−24 J

Quizplus · Accepted Answer

We can use the conservation of energy principle for this problem.

Initial mechanical energy at point A: 
Ei = mgh + 1/2 * mv^2 = 1.2 * 9.81 * 2 + 1/2 * 1.2 * (3.2)^2 = 28.2472 J (potential energy + kinetic energy)

Final mechanical energy at point B (due to friction force): 
Ef = mgh + 1/2 * mv^2 = 1.2 * 9.81 * 2 + 1/2 * 1.2 * (6.0)^2 = 36.4472 J (potential energy + kinetic energy)

Change in mechanical energy: 
ΔE = Ef - Ei = 36.4472 - 28.2472 = 8.2 J

However, the problem asks for the change in mechanical energy done on the system by the force of friction, which is opposite to the direction of motion. Therefore, we need to put a negative sign in front of ΔE.

So, the final answer is:

ΔE = −8.2 J ≈ −8.1 J (rounded to one decimal place)

Therefore, the best choice is C.

A 12-Kg Mass Is Projected Down a Rough Circular Track (Radius