A force F required to keep a 5 kg mass moving round a cycle of radius 3.5 m at a speed of 7 ms-1. What is the speed, if the force is tripled?
A
4.0 ms-1
B
6.6 ms-1
C
12.1 ms-1
D
21.0 ms-1
Verified Correct Answer: C.
Physical Principles & Numerical Solution
The force which is centripetal
F = \(\frac{MV^2}{R}\)
= \(\frac{5 \times 7^2}{3.5}\)
= 70
Since force is 70 N when speed is 7m/s
Speed = ? when force 70N is tripled → 70 * 3 = 210N
210 = \(\frac{5\timesv^2}{3.5}\)
210 \times 3.5 = 5 * v\(^2\)
735 \div 5 = v\(^2\)
v\(^2\) = 147
v =12.12m/s
Source: WAEC / JAMB / NECO Classroom Archive (Record #14484)