The question i sent to you

The normal reaction force $N$ on body P is equal to its weight, as it rests on a horizontal surface.

$$N = m_P g$$

Substitute the given values:

$$N = (10)(10)$$ $$N = \boxed{\text{100 N}}$$

The frictional force $F_f$ is given by the product of the coefficient of friction and the normal reaction force.

$$F_f = \mu N$$

Substitute the values:

$$F_f = (0.5)(100)$$ $$F_f = \boxed{\text{50 N}}$$

The net force causing acceleration is the weight of Q minus the frictional force on P. The total mass of the system is $m_P + m_Q$.

$$a = \frac{m_Q g - F_f}{m_P + m_Q}$$

Substitute the values:

$$a = \frac{(15)(10) - 50}{10 + 15}$$ $$a = \frac{150 - 50}{25}$$ $$a = \frac{100}{25}$$ $$a = \boxed{\text{4 m/s}^2}$$

Consider the forces acting on body Q. The weight of Q acts downwards, and the tension acts upwards. The net force causes Q to accelerate downwards.

$$m_Q g - T = m_Q a$$

Rearrange the equation to solve for $T$:

$$T = m_Q g - m_Q a$$ $$T = m_Q (g - a)$$

Substitute the values:

$$T = (15)(10 - 4)$$ $$T = (15)(6)$$ $$T = \boxed{\text{90 N}}$$

This is the net force on the entire system, which is the product of the total mass and the common acceleration.

$$F_{\text{net}} = (m_P + m_Q) a$$

Substitute the values:

$$F_{\text{net}} = (10 + 15)(4)$$ $$F_{\text{net}} = (25)(4)$$ $$F_{\text{net}} = \boxed{\text{100 N}}$$