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