Question

1. A 1 kg block of wood is moving with a velocity of 10 m s-1 on top of a table. The coefficient of kinetic friction between the block of wood and the table is 0.1. The block of wood is a cube with sides of 12 cm.

a) How far does the block move before coming to a stop?
b) If momentum id conserved, where did the momentum of the block go?
c) The block slows down because of the frictional force between the block and the block. However, if the frictional force acts at the bottom of the block and the blocks center of mass is in the center of the block then there must be a torque acting on the block. Calculate this torque. Why doesn't this torque cause the block to rotate?

Answer 1

Given that :

mass of the block of wood, m = 1 kg

initial velocity of the block of wood, v_o = 10 m/s

coefficient of kinetic friction between the block of wood and the table, _{$\mu$ k} = 0.1

side of block, b = 12 cm = 0.12 m

(a) the block moves before coming to a stop at a distance which is given as ::

acceleration due to friction, a = _{$\mu$ k} g                                                     { eq.1 }

where, g = - 9.8 m/s²

inserting the values in eq.1,

a = (0.1) (- 9.8 m/s²)

a = - 0.98 m/s²

now, using equation of motion 1 -

v_f² - v_o² = 2 a s { eq.2 }

where, v_f = final velocity = 0 m/s

inserting the values in eq.2,

(0 m/s)² - (10 m/s)² = 2 (-0.98 m/s²) s

s = (100 / 1.96) m

s = 51 m

(b) If momentum is conserved, then it means that "there is no external force in the form of friction which is acting on the body''.

(c) The block slows down because of the frictional force between the block and the block.

the net torque acting on the block of wood is zero because due to friction, torque is balalnced by the torque due to the weight.

Torque doesn't cause the block to rotate because the normal force of the surface on the block supplies a counter-balancing torque. due to this torque, the system is static and the block will not rotate.