A man lifts a 1.0 kg stone vertically with his hand at a constant upward acceleration of 1.8m/2^2. what is the magnitude of the total force of the stone on the man's hand?

In Newtons downward?

Well, if the stone is being lifted upward, I don't think there would be a force downward, unless gravity suddenly decided to take a nap. So, in this case, the magnitude of the total force of the stone on the man's hand would be equal to the weight of the stone, which is approximately 9.8 newtons downward, due to gravity. Keep in mind that the man's hand has to exert a force slightly greater than this to lift the stone with an upward acceleration. So, the stone may be a burden, but at least gravity doesn't add insult to injury by pulling it down even harder!

To find the magnitude of the total force of the stone on the man's hand, we need to consider Newton's second law of motion, which states that force is equal to mass multiplied by acceleration.

The mass of the stone is given as 1.0 kg, and the acceleration is given as 1.8 m/s^2. Therefore, the force exerted by the stone on the man's hand can be calculated as follows:

Force = mass * acceleration
Force = 1.0 kg * 1.8 m/s^2
Force = 1.8 N

So, the magnitude of the total force of the stone on the man's hand is 1.8 Newtons.

Now, you also asked for the force in Newtons downward. Since the problem states that the stone is lifted vertically with a constant upward acceleration, the force exerted by the stone on the man's hand is directed upward. Therefore, the force cannot be measured as Newtons downward.

The man's hand pushes the stone upward with force F = m a + m g (since F - Mg is the net force)

The stone pushes the man's hand downward with the same force. (Because of Newton's third law)