A bodybuilder pulls a 60 kg tyre with a force of 250 N at an angle of 20 degrees to the horizontal. If he pulls for 25 m, calculate the work done by the bodybuilder and the work done against friction,

The work done by a force at an angle is given by $W = F d \cos \theta$.
Here, $F = 250 \, \text{N}$, $d = 25 \, \text{m}$, and $\theta = 20^\circ$.

$$W_B = F d \cos \theta$$ $$W_B = (250 \, \text{N})(25 \, \text{m}) \cos(20^\circ)$$ $$W_B = 6250 \, \text{J} \times 0.9397$$ $$W_B \approx \boxed{5873.125 \, \text{J}}$$

2- Provided that the tyre was initially at rest and there is an additional frictional force of $220 \, \text{N}$:

The work done against friction is $W_f = F_f d$.
Here, $F_f = 220 \, \text{N}$ and $d = 25 \, \text{m}$.

$$W_f = F_f d$$ $$W_f = (220 \, \text{N})(25 \, \text{m})$$ $$W_f = \boxed{5500 \, \text{J}}$$

The net work done on the tyre is the work done by the bodybuilder minus the work done against friction. This net work equals the change in kinetic energy ($\Delta K_E$). Since the tyre starts from rest, the initial kinetic energy is 0.

$$W_{net} = W_B - W_f$$ $$W_{net} = 5873.125 \, \text{J} - 5500 \, \text{J}$$ $$W_{net} = 373.125 \, \text{J}$$

So, the kinetic energy provided to the tyre is:

$$K_E = \boxed{373.125 \, \text{J}}$$

The kinetic energy is given by $K_E = \frac{1}{2}mv^2$. We are given $K_E = 373 \, \text{J}$ and $m = 60 \, \text{kg}$.

$$K_E = \frac{1}{2}mv^2$$ $$373 \, \text{J} = \frac{1}{2}(60 \, \text{kg})v^2$$ $$373 \, \text{J} = (30 \, \text{kg})v^2$$ $$v^2 = \frac{373 \, \text{J}}{30 \, \text{kg}}$$ $$v^2 = 12.433 \, \text{m}^2/\text{s}^2$$ $$v = \sqrt{12.433 \, \text{m}^2/\text{s}^2}$$ $$v \approx \boxed{3.526 \, \text{m/s}}$$