In a second ODI match between England and India Bhuwnesh kumar bowled his medium fast bowling. He starts his run up from the distance of `50m` from the stumps (bowling end) and at the time of bowling a ball,his speed is `36(km)//(hr)` and the speed of ball is `144(km)//(hr)`.
If his mass is `70 kg` and mass of the ball is `(1)/(4)kg` and the heat produce in his body is one tenth of the total work done by him, then
If he accelerates uniformly to the sumps find the average friction force acting on him `:-`

Options

A. `750` N
B. `70.25` N
C. `90.25` N
D. `70` N

Explanation

To solve the problem step by step, we will follow the principles of physics related to work, energy, and friction.

### Step 1: Convert Speeds to Meters per Second
First, we need to convert the speeds from kilometers per hour to meters per second.

- Bhuvnesh's speed:
\[ 36 \text{ km/hr} = \frac{36 \times 1000}{3600} = 10 \text{ m/s} \]

- Ball's speed:
\[ 144 \text{ km/hr} = \frac{144 \times 1000}{3600} = 40 \text{ m/s} \]

### Step 2: Calculate the Change in Kinetic Energy
Next, we need to calculate the change in kinetic energy of Bhuvnesh as he runs towards the stumps.

- Initial kinetic energy (KE_initial) = 0 (he starts from rest)
- Final kinetic energy (KE_final) = \(\frac{1}{2} m v^2\)
where \(m\) is the total mass (Bhuvnesh + ball) and \(v\) is his final speed.

- Total mass:
\[ m = 70 \text{ kg} + \frac{1}{4} \text{ kg} = 70.25 \text{ kg} \]

- Final kinetic energy:
\[ KE_{final} = \frac{1}{2} \times 70.25 \times (10)^2 = \frac{1}{2} \times 70.25 \times 100 = 3512.5 \text{ J} \]

### Step 3: Calculate the Work Done by Friction
According to the work-energy principle, the work done by the friction force is equal to the change in kinetic energy.

- Work done by friction (W) = Change in KE = KE_final - KE_initial = \(3512.5 \text{ J} - 0 = 3512.5 \text{ J}\)

### Step 4: Relate Work Done to Friction Force
The work done by the friction force can also be expressed as:
\[ W = F_{friction} \times d \]
where \(d\) is the distance traveled (50 m).

Rearranging gives:
\[ F_{friction} = \frac{W}{d} = \frac{3512.5}{50} = 70.25 \text{ N} \]

### Step 5: Conclusion
The average friction force acting on Bhuvnesh is:
\[ F_{friction} = 70.25 \text{ N} \]