## 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 **

### 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, F_{x3} = ?

Speed when we triple the force, V_{x3} = ?

**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 = mV^{2} / r

We will then multiply the result from the value of centripetal force by 3 to obtain F_{x3}

**Step 2: **We will use the formula F_{x3} = mV_{x3}^{2} / r to make the speed subject of the formula

V_{x3} = √(F_{x3} 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 = mV^{2} / r

F = mV^{2} / r = (5 x 7^{2}) / 3.5 = 70 N

After tripling the force, F_{x3} = 3 x 70 = 210 N

**Step 2:** Similarly, we will equally apply the formula V_{x3} = √(F_{x3} r / m) to find the speed

V_{x3} = √(F_{x3} r / m) = √(210 x 3.5 / 5) = √147

Hence,

The speed V_{x3} = 12.1 m/s

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

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