Question

a block of mass m=2.0 kg, initially at h=1.5 m, is released with an initial speed of 10 m/s and slides along a track. the track is frictionless except from point A to point B, where the length L is 3.0 m and the coefficient of kinetic friction is Uk= 0.3. Note: The block can be treated as a point object. the gravitational energy potential energy on the ground level is taken to be zero

1. what is the intial gravitational energy og the block? 2.what is the speed of the block at point A? 3. how much energy is dissipated as thermal energy ? 4. what is the speed of the block after passing the point B ?

Answer 1

1) the intial gravitational energy og the block = m*g*h

= 2*9.8*1.5

= 29.4 J

2) use conservation of energy

final KE = initial potential energy + kinetic energy

(1/2)*m*vf^2 = m*g*h + (1/2)*m*vi^2

vf^2 = 2*g*h + vi^2

vf = sqrt(2*g*h + vi^2)

= sqrt(2*9.8*1.5 + 10^2)

= 11.4 m/s

3) the energy is dissipated as thermal energy = -workdone by friction

= -fk*d*cos(180)

= fk*d

= mue_k*m*g*d

= 0.3*2*9.8*3

= 17.6 J

4) acceleration of the block on rough surface, a = -g*mue_k

= -9.8*0.3

= -2.94 m/s^2

now use, vf^2 - vi^2 = 2*a*L

vf = sqrt(vi^2 + 2*a*L)

= sqrt(11.4^2 + 2*(-2.94)*3)

= 10.6 m/s