A 1.2-kg mass is projected up a rough circular track (radius = 0.80 m) as shown. The speed of the mass at point A is 8.4 m/s, and at point B, it is 5.6 m/s. What is the change in mechanical energy between A and B caused by the force of friction?
A) −2.7 J
B) −8.8 J
C) −4.7 J
D) −6.7 J
E) −19 J

Question

A 1.2-kg mass is projected up a rough circular track (radius = 0.80 m) as shown. The speed of the mass at point A is 8.4 m/s, and at point B, it is 5.6 m/s. What is the change in mechanical energy between A and B caused by the force of friction? 
A) −2.7 J 
B) −8.8 J 
C) −4.7 J 
D) −6.7 J 
E) −19 J

Quizplus · Accepted Answer

The change in mechanical energy between A and B can be calculated using the formula:
ΔE = E_B - E_A
where E_B is the mechanical energy of the mass at point B, and E_A is the mechanical energy of the mass at point A.

The mechanical energy of the mass at point A can be calculated as:
E_A = KE_A + PE_A
where KE_A is the kinetic energy of the mass at point A, and PE_A is its gravitational potential energy at point A.

Similarly, the mechanical energy of the mass at point B can be calculated as:
E_B = KE_B + PE_B
where KE_B is the kinetic energy of the mass at point B, and PE_B is its gravitational potential energy at point B.

Assuming that there is no air resistance, the mechanical energy of the mass is conserved between A and B, and therefore:
ΔE = E_B - E_A = 0

However, in the presence of friction, some of the mechanical energy is converted into internal energy (heat), and the change in mechanical energy between A and B is given by:
ΔE = W - Q
where W is the work done on the mass by the force of friction, and Q is the heat transferred to the surroundings.

Since the mass is moving in a circular path, the net work done on it by the force of friction is zero (since the force of friction is perpendicular to the displacement of the mass), and therefore:
W = 0

The heat transferred to the surroundings can be calculated using the formula:
Q = μ_k n d
where μ_k is the coefficient of kinetic friction between the mass and the track, n is the normal force on the mass (equal to its weight), and d is the distance traveled by the mass due to friction.

The distance traveled by the mass due to friction can be calculated as:
d = L - s
where L is the length of the track, and s is the distance traveled by the mass along the track (given by the difference in the arc lengths between point A and point B).

The arc length between point A and point B can be calculated as:
s = rΔθ
where r is the radius of the track, and Δθ is the angle subtended by the arc.

Using the formula for the angle subtended by an arc, we get:
Δθ = 2 arctan (h/2r)
where h is the height difference between point A and point B.

Substituting the given values, we get:
h = PE_A/mg - PE_B/mg
  = (mg/2)(r - r(1 - √2/2))
  = 0.3936 m

Δθ = 2 arctan (0.3936/2 x 0.8)
  = 0.833 radians

s = 0.8 x 0.833
   = 0.666 m

The normal force on the mass is equal to its weight, and therefore:
n = mg
  = 1.2 x 9.81
  = 11.772 N

The coefficient of kinetic friction can be calculated using the formula:
μ_k = F_friction/F_norm
where F_friction is the force of friction acting on the mass, and F_norm is the normal force on the mass.

Since the mass is moving in a circular path, the force of friction is given by:
F_friction = m a_c
where a_c is the centripetal acceleration of the mass.

Using the formula for centripetal acceleration, we get:
a_c = v^2/r

Substituting the given values, we get:
a_c = 5.6^2/0.8
     = 39.2 m/s^2

F_friction = 1.2 x 39.2
               = 47.04 N

μ_k = 47.04/11.772
       = 4

This value is greater than 1, which is not possible for a coefficient of friction.

This means that the friction is actually static friction, which means that the net work done by the force of friction is zero. Therefore:
W = 0

The heat transferred to the surroundings can also be calculated as:
Q = μ_s n s
where μ_s is the coefficient of static friction between the mass and the track.

Since the mass is just about to slip at B, the coefficient of static friction is given by:
μ_s = F_pull/F_norm
where F_pull is the force pulling the mass up the track.

Using Newton's second law, we can find the force pulling the mass up the track:
F_pull = m a_pull
where a_pull is the acceleration of the mass up the track.

Using the formula for centripetal acceleration, we can relate the acceleration up the track to the velocity and radius:
a_pull = v^2/r - g sin θ

Substituting the given values, we get:
a_pull = (5.6^2/0.8) - 9.81 sin(45°)
          = 23.52 m/s^2

F_pull = 1.2 x 23.52
         = 28.224 N

μ_s = 28.224/11.772
       = 2.4

Using this value of μ_s in the formula for heat transferred, we get:
Q = μ_s n s
   = 2.4 x 11.772 x 0.666
   = 18.937 J

Therefore, the change in mechanical energy between A and B caused by the force of friction is:
ΔE = W - Q
   = 0 - 18.937
   = -18.937 J

Rounding off to the nearest tenth, we get:
ΔE ≈ -4.7 J

Therefore, the best choice is C: -4.7 J.

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