Question

a 0.426-kg block sliding from A to B along a frictionless surface. When the block reaches B, it continues to slide along the horizontal surface BC where the kinetic frictional force acts. As a result, the block slows down, coming to rest at C. The kinetic energy of the block at A is 39.0 J, and the heights of A and B are 12.4 and 6.10 m above the ground, respectively. (a) What is the value of the kinetic energy of the block when it reaches B? (b) How much work does the kinetic frictional force do during the BC segment of the trip?

Answer 1

given that :

kinetic energy of the block at A, K.E_A = 39 J

mass of the block, m = 0.426 kg

height of A above the ground, h_A = 12.4 m

height of B above the ground, h_B = 6.1 m

(a) the value of the kinetic energy of the block when it reaches B which is given as :

using a conservation of energy, the total energy at B which equals to at A.

K.E_B + P.E_B = K.E_A + P.E_A                  { eq. 1 }

K.E_B = K.E_A + P.E_A - P.E_B

P.E = potential energy = mgh

inserting the value of 'P.E' in above eq.

K.E_B = K.E_A + mg (h_A - h_B) { eq. 2 }

inserting the given values in eq.2,

K.E_B = (39 J) + (0.426 kg) (9.8 m/s²) [(12.4 m) - (6.1 m)]

K.E_B = (39 J) + (4.17 kg.m/s²) (6.3 m)

K.E_B = 65.3 J

(b) during the BC segment of the trip.

using a work-energy theorem, the total final mechanical energy minus total initial mechanical energy which is given as :

W_f = (K.E_C + P.E_C) - (K.E_B + P.E_B) { eq.3 }

W_f = (K.E_C - K.E_B) + mg (h_C - h_B)

where, K.E_C = kinetic energy ofblock at C = 0

h_C = height above the ground = 6.1 m

inserting the values in above eq.

W_f = 0 - K.E_B + (0.426 kg) (9.8 m/s²) [(6.1 m) - (6.1 m)]

W_f = - K.E_B + 0

W_f = - K.E_B                          { eq. 4 }

inserting the value of 'K.E_B' in eq.4,

W_f = - (65.3 J)

work done by the frictional force is negative, because this force points opposite to the displacement of the block.