(i)

W CL + W f = KE W CL = KE – W f

During accelerated motion negative work is done against friction and there is also change is

kinetic energy. Hence network needed is +ve.

During uniform motion work is done against friction only and that is +ve.

During retarded motion, the load has to be stopped in exactly 50 metres. If only friction is

considered then the load stops in 12.5 metres which is less than where it has to stop.

Hence the camel has to apply some force so that the load stops in 50m (>12.5 m). Therefore

the work done in this case is also +ve.0

(ii) W CL | accelerated motion = KE – W friction

where W CL is work done by camel on load.

=

= =

similarly, W CL | retardation = KE – W friction

– [µ k mg.50] =

∴

= = ⇒ 5 : 3

(iii) Maximum power = F max × V

Maximum force applied by camel is during the accelerated motion.

We have V 2 – U 2 = 2as

25 = 0 2 + 2.a.50

a = 0.25 m/s 2 ; for accelerated motion

F C – f = ma

F C = µmg + ma = 0.1×1000×10+1000×2.5

= 1000 + 250 = 1250 N

This is the critical point just before the point where it attains maximum velocity of almost 5 m/s.

Hence maximum power at this point is = 1250 × 5 = 6250 J/s.