Block of mass m are released from rest and they slide down the inclined surface as shown in figure. Show that the work done by the frictional forces on the blocks are same in all the cases. Also calculate the speeds with which the blocks reach point B. The coefficient of friction for all the surface is `mu_k`.

Explanation

For figure (a)
Frictional force on the block `=mu_kN=mu_kmgcostheta`
If S be the length of the incline, work done by the frictional force on the block when the block slides down from A to B,

`W_(fl)=(mu_kmgcostheta)(s)`
{Force and displacement are antiparallel}
`=-mu_kmg(s cos theta)=-mu_kmg`
Figure (b)
Frictional forces on the block at the plans with the inclination `theta_1` and `theta_2` are `mu_kmgcostheta_1` and `mu_kcostheta_2`. If the lengths of the planes be `S_1` and `S_2`, work done by fricitional force on the block,

`W_(fk)=-(mu_kmgcostheta_1)s_1-(mu_kmgcostheta_2)s_2`
`=-mu_kmg(s_1costheta_1+s_2costheta_2)=-mu_kmgl`
For Figure (c)
Let the length of the inclines with inclination `theta_1`, `theta_2`, and `theta_3` be `s_1`, `s_2`, and `s_3`, respectilvey. At these planes, the frictional force on the block are `mu_kmgcostheta_1`, `mu_kmgcostheta_2`, and `mu_kmgcostheta_3`, respectively. Work done by the frictional force on the block,

`W_(fk)=-(mu_kmgcostheta_1)s_1-(mu_kmgcostheta_2)s_2-(mu_kmgcostheta_3)s`
`=-mu_kmg(s_1costheta_1+s_2costheta_2+s_3theta_3)=-mu_kmgl`.
For Figure (d)
Let the block be at point C where the tangent to the parabola makes an angle `theta` with the horizontal as shown in figure.
Frictional force on the block at
`C=mu_kmgcostheta`, as shown. Consider an infinitesimal displacement `ds` at C.
Work done by the frictional force over `ds`,

`dW=mu_1mgcosthetads`
Work done from A to B,
`W_(fk)=int_A^BdWint_A^B-mu_kmgcosthetads`
`=-int_A^Bmu_kmgdx=-mu_kmgl`
Calculations show that the work done by the frictional force in each case are same.
Speed at B:
Using work-energy theorem.
`W_(mg)=W_(fk)=K_2-K_1`
`impliesmgh-m_kmgl=1/2mv^2=0`
(Block is at rest at A, `K_1=0`)
`implies 2g(h-m_kl)=v^2impliesv=sqrt(2g(h-mu_kl))`
In each case, the block will reach B with same speed, which equals `sqrt(2g(h-mu_kl))`.