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A force F required to keep a 5 kg mass moving round a cycle of radius 3.5 m at a speed

Question

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?

Answer

The final answer to this question (the speed, when we triple the force) is 12.1 m/s

A force f is required to keep a 5kg mass
A force f is required to keep a 5 kg mass

Explanation

Data: Unrevealed information from the question

Mass of the object, m = 5 kg

Radius of the circle, r = 3.5 m

Velocity of the object, V = 7 m/s

Unknown: Unrevealed information from the question

Force towards the center (centripetal force), F = ?

When we triple the force, Fx3 = ?

Speed when we triple the force, Vx3 = ?

Formula: The equation that will help us solve the problem

Step 1: To find the force towards the center, we will apply the formula F = mV2 / r

We will then multiply the result from the value of centripetal force by 3 to obtain Fx3

Step 2: We will use the formula Fx3 = mVx32 / r to make the speed subject of the formula

Vx3 = √(Fx3 r / m) [We will apply this formula to find the speed after tripling our force]

Solution

Step 1: We will insert our data into the formula F = mV2 / r

F = mV2 / r = (5 x 72) / 3.5 = 70 N

After tripling the force, Fx3 = 3 x 70 = 210 N

Step 2: Similarly, we will equally apply the formula Vx3 = √(Fx3 r / m) to find the speed

Vx3 = √(Fx3 r / m) = √(210 x 3.5 / 5) = √147

Hence,

The speed Vx3 = 12.1 m/s

Therefore, after tripling the force. The speed will become 12.1 meters per second.

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