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Monday, December 25, 2017

A truck of mass 500 kg moves from rest at the top of a section of track 400 m long and 30 m high, as shown. The frictional force acting on the truck is 250 N throughout its journey.



Question 6
A truck of mass 500 kg moves from rest at the top of a section of track 400 m long and 30 m high, as shown. The frictional force acting on the truck is 250 N throughout its journey.


What is the final speed of the truck?
A 14 m s-1                         B 24 m s-1                         C 31 m s-1                         D 190 m s-1





Reference: Past Exam Paper – November 2016 Paper 11 & 13 Q19





Solution:
Answer: A.

From the conservation of energy,
GPE of truck at top = KE of truck at bottom + Work done against friction


GPE of truck at top = mgh = 500 × 9.81 × 30 = 147 150 J


Work done against friction = F × d = 250 × 400 = 100 000 J


GPE of truck at top = KE of truck at bottom + Work done against friction
147 150 = KE of truck at bottom + 100 000

KE of truck at bottom = 147 150 – 100 000 = 47 150 J
½ mv2 = 47 150

Final speed v = (2×47150 / 500) = 13.7 m s-1 ≈ 14 m s-1

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