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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