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?
4.0 ms-1
6.6 ms-1
12.1 ms-1
21.0 ms-1
Explanation
Video Explanation
Post your Contribution
Discussions (16)
The answer is wrong because F=mv^2/r
so F=5×7^2/3.5
F=70
when the force is tripled
F=70×3=210
210=5×v^2/3.5
210×3.5=5v^2
735÷5=v^2
v^2=147
v=12.1
The answer is wrong
F=mv^2/r
F=5*7^2/3.5
=70N
when force is trippled=70*3=210
substituting into the same formula,
210=5*v^2/3.5
make v subject of formula
v^2=210*3.5/5
v=SQR147
v=12.1m/s
THE ANSWER IS CORRECT, OPTION C JUST THAT THE PROCESS IS NOT CLEAR, WALK WITH ME
F=MV^2-------[1]
Keep M&R CONSTANT
FV^2
F1/V1^2==F2/V2^2
F2=3F1[THE FORCE TRIPLED]
F1/V1^2=3F1/V2^2
F1/7^2=3F1/V2^2
F1/49=3F1 /V2 ^2
CROSS MULTIPLYING
V2^2 =147
V2=SQUARE ROOT OF 147
V2=12.1
THAT'S ALL
F =;[mv²]/r
F = 70
when F is tripled
70x3 =[ mv²]/r
210 = [mv²]/r
make v subject of formula
v = sqrt[(210 • r)/m]
v = sqrt[147]
v = 12.124
F=mv^2/r
F= 5*7^2/3.5
F= 70
when force is tripled
F =3F
F = 3*70
F=210
Substituting into the formula
210 = 5*V^2/3.5
Cross multiply to make v the subject of the formula
210*3.5 = 5V^2
Following mathematical processess
V = 12.1 or 7✓3
