# Question Video: Mass, Weight, and Applied Force Physics • 9th Grade

A rock with a mass of 2.5 kg is thrown vertically upward by an applied force of 60 N. What is the rock’s weight? What is the net vertically upward force applied to the rock? What is the rock’s rate of upward vertical acceleration just before it is released? What is the rock’s rate of upward vertical acceleration just after it is released?

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### Video Transcript

A rock with a mass of 2.5 kilograms is thrown vertically upward by an applied force of 60 newtons. What is the rock’s weight? What is the net vertically upward force applied to the rock? What is the rock’s rate of upward vertical acceleration just before it is released? What is the rock’s rate of upward vertical acceleration just after it is released?

Okay, so in this question, we’ve been told that we have a rock that is thrown vertically upward. We’ve been told that the rock is thrown vertically upward by an applied force of 60 newtons. So we can say that the hand, for example, doing the throwing of the rock is applying an upward force of 60 newtons. But as well as this, we need to recall that the rock is going to have a gravitational force acting on it because the rock is found in the gravitational field of the Earth.

Therefore, the rock is going to have a weight force, which we’ll call 𝑊, acting on it. And the first part of the question actually asks us to find the rock’s weight. So to do this, we can recall that the weight of an object is found by multiplying the mass of that object by the gravitational field strength of the Earth in this case. And from the question, we already know the mass of the rock. And we can recall that the gravitational field strength of the Earth is 9.8 metres per second squared.

Therefore, to find the weight of the rock 𝑊, we’ll say that this is equal to 2.5 kilograms — that’s the mass of the rock — multiplied by 9.8 metres per second squared, the gravitational field strength. Now before we evaluate this, we can see that we’re working in base units, kilograms for mass and metres per second square for the gravitational field strength, otherwise known as the acceleration due to gravity. Therefore, the weight of the rock is going to be in its own base unit, which is the newton. And this is because weight is a force.

And so the right-hand side of the equation becomes 24.5 newtons. Hence, we can say that the weight of the rock, the answer to the first part of the question, is 24.5 newtons.

Moving on then, we need to find the net vertically upward force applied to the rock. In other words, we need to find the overall force acting on the rock when the two individual forces acting on the rock are the 60-newton upward force applied by the hand and the downward gravitational force, otherwise known as the weight. And because in the question we’ve been asked to find the net vertically upward force, let’s arbitrarily choose the upward direction to be positive and the downward direction therefore will be negative.

Now it’s of course worth noting that we’re trying to find the net vertically upward force applied to the rock just before the rock has been let go from the hand. Because once the hand lets go of the rock, this 60-newton upward force will no longer be acting on the rock. However, because we’ve been asked to find the net force, we therefore need to consider the two forces acting on the rock. And this is just before the rock has been released.

Hence, we can say that the net force on the rock before it’s been released, which we’ll call 𝐹 subscript net, is equal to the upward 60-newton force, which is positive because it’s pointing in the upward direction, minus its weight, which we’ve seen is 24.5 newtons, because the weight is acting in the downward direction. And so the overall or resultant or net upward force is 35.5 newtons. And this is our answer to the second part of the question. So let’s move on to the third part.

What is the rock’s rate of upward vertical acceleration just before it is released?

Well, like we said already, just before the rock is released from the hand, it has two forces acting on it: the upward 60-newton force and the downward weight force. And as we’ve already seen, the net force on the rock because of this is 35.5 newtons upward. And that’s because this net force is positive and we said the positive direction was the upward direction.

Then we can recall Newton’s second law of motion, which tells us that the net force on an object is equal to the object’s mass multiplied by the acceleration it experiences. Now in this scenario, just before the rock is released from the hand, we already know the net force and we also know the mass of the rock. Therefore, we can solve for the acceleration of the rock. And coincidentally, because the net force on the rock is in the vertically upward direction, this means that using our equation will give us the rock’s rate of upward vertical acceleration, just as we need.

So to solve for the acceleration, let’s divide both sides of this equation by the mass of the rock 𝑚. Doing this means that the 𝑚 on the right-hand side cancels. And what we’re left with is that the net force on the rock divided by the mass of the rock is equal to the acceleration experienced by the rock.

Then when we plug in our values, we see that the acceleration experienced by the rock is equal to the net force, 35.5 newtons, divided by the mass, 2.5 kilograms. This evaluates to an acceleration of 14.2 metres per second squared. And therefore, we found the answer to the third part of our question. So we can move on to the final part.

What is the rock’s rate of upward vertical acceleration just after it is released?

Well, just after the rock is released, the hand is no longer exerting that upward 60-newton force on the rock. Therefore, the only force now acting on the rock is the weight of the rock. And so the new net force acting on the rock is simply the weight of the rock.

Writing this mathematically, we can say that the net force on the rock, 𝐹 subscript net, comma after the rock has been released from the hand is now simply given by the weight of the rock. But remember, the weight of the rock is acting in the downward or negative direction. Therefore, we can say that the net force on the rock is equal to negative 𝑊.

But then we know that the weight of an object is defined as the mass of that object multiplied by the gravitational field strength. Therefore, using that equation, we can say that negative 𝑊 is equal to negative the mass of the rock multiplied by the gravitational field strength of Earth. And we also know that the net force on an object at any time is given by multiplying the mass of that object by the acceleration experienced by the object. And the effect of all of this is to realise the mass of the rock multiplied by its acceleration is the same thing as negative the mass of the rock multiplied by the gravitational field strength of Earth.

Therefore, we can get rid of anything else in the middle of this equation. And we can say that 𝑚𝑎 is equal to negative 𝑚𝑔. Then we can divide both sides of the equation by the mass of the rock, 𝑚. And this results in 𝑚 cancelling on both sides. And so what we’re left with is that the acceleration experienced by the rock just after the rock is released from the hand is equal to negative 𝑔 or negative 9.8 metres per second squared. And that is our answer to the very final part of this question.

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