A 60 kg mass is accelerated vertically upwards at 10 m/s2. What is the magnitude of the applied force? (g = 10 m/s2)

There are two main forces:

• The weight of the mass ($W$) acting downwards.
• The applied force ($F_{app}$) acting upwards.

The weight is given by $W = mg$.
Given mass $m = 60 \, \text{kg}$ and acceleration due to gravity $g = 10 \, \text{ms}^{-2}$.

$$W = (60 \, \text{kg}) \times (10 \, \text{ms}^{-2}) = 600 \, \text{N}$$

Newton's Second Law states that the net force ($F_{net}$) acting on an object is equal to its mass ($m$) times its acceleration ($a$).
Since the acceleration is vertically upwards, the net force must also be upwards.

$$F_{net} = F_{app} - W$$ $$ma = F_{app} - W$$

Rearrange the equation to find $F_{app}$:

$$F_{app} = ma + W$$

Substitute the given values: $m = 60 \, \text{kg}$, $a = 10 \, \text{ms}^{-2}$, and $W = 600 \, \text{N}$.

$$F_{app} = (60 \, \text{kg}) \times (10 \, \text{ms}^{-2}) + 600 \, \text{N}$$ $$F_{app} = 600 \, \text{N} + 600 \, \text{N}$$ $$F_{app} = 1200 \, \text{N}$$

The calculated force is $1200 \, \text{N}$.

(a) 300N

(b) 600N

(c) 1200N

(d) 400N

The correct option is (c).

The final answer is $\boxed{\text{1200N}}$.