Suppose a 65-kg person jumps from a height of 3.0 m down to th...

Question

Suppose a 65-kg person jumps from a height of 3.0 m down to the ground.

(c) Estimate the decelerating force if the person lands stiff-legged so d ≈ 1.0 cm .

Accepted Answer

Welcome back. Everyone in this problem. A climber accidentally slips off a cliff and falls freely from a height of 20 m before landing in a snow bank that compresses 1.5 m to cushion the fall. If the climber has a mass of 60 kg, calculate the average decelerating force exerted by the snow bank during the climber's landing. Here, we have a diagram of the climber falling from 20 m. And for our answer choices, A says that the average decelerating force is 3.9 multiplied by 10 cubed Newtons. B says it's 5.9 multiplied by 10 cubed newtons C 7.8 multiplied by 10 cubed newtons and D 9.8 multiplied by 10 cubed newtons. Now in our diagram, OK. Let's make sure we understand what we're trying to figure out here. I think I'll use this, this first representation. OK. So we know that the climber falls, OK? And he falls into a snow bank that compresses 1.5 m to cushion the floor. So we know that here on our diagram, this distance is 1.5 m. OK? And we are trying to figure out the average decelerating force exerted by the snow bank during the, the climbers landing. In other words, when the, when the, when the, when the climber gets to the snow bank, ok, when the climber eventually gets to the snow bank, oops, let me fix that a bit. And let's just use, let's just use some dotted lines here to show that we're not there yet. It's just our representation. Ok? Now, when the clamor gets to that point, OK, then we want to find a force that's exerted by the snow bank. In other words, we want to find a force, let's call it F OK. That's exerted by the snow bank. I think I need to move this here as well. So the question is what is going to be that force if that is exerted on that is exerted on the climber during landing? What do we know about for? Well, recall that by Newton's second law of motion force is directly proportional to acceleration by the formula force equals mass multiplied by acceleration. No, here we have the mass of the climber, but we don't have the acceleration. So we'll need to figure that out in order to find the force. Now, if we're talking acceleration, remember acceleration is defined as the change in velocity over time. So at the very least, we'll need to figure out the change in velocity right before impact and right after impact. So let's see if we can figure that out. OK. So we know when it comes to our snow bank after impact, the the climber is stopped. Ok. He lands in the snow bank, it cushions his fall. His velocity is going to be zero. But we don't know the velocity just before impact. That's what we need to figure out. So to find that we need to find the velocity right, as the climber is about to enter the snow bank. So let's call that velocity vo now, do we know anything that relates velocity, initial velocity and final velocity for a period of freefall here for our distance of 20 m? Well, we also know that let me put that in blue here. We also know that our final velocity squared V squared equals our initial velocity U squared plus two A that relates the distance as to the acceleration and the initial and final velocities here in free fall. Make a note here in freefall. OK. Our final velocity for this problem is going to be V knot. So that's VN squared, our initial velocity U is going to be zero because he's coming from the cliff, it's from a standstill and the acceleration is going to be the acceleration due to gravity. G. And the distance is is going to be the height that the climber felt in this case, as we said U squared is zero. So that means our velocity just before impact squared is going to be equal to two GH which means that if we square root both sides our velocity just before impact is going to be the square root of two G where our height is 20 m. So we can use that then to solve for V. Not OK. Because no, that means it's going to be equal to the square root of two multiplied by the acceleration due to gravity. We take to be 9.8 m per second squared multiplied by our height, which is 20 m. So let's find the square root of that expression. Now, when we do that, we should get our velocity to be equal to 19.8 m per second. So what that means is that at right at the point where our climber hits the snow bank, his velocity is 19.8 m per second. Now, if we're going to find acceleration at this point, we have our initial velocity, the velocity just before impact, we know that our velocity after impact is zero. And we know that the snow bank is 1.5 m high. So the climber would travel a distance of 1.5 m when you think about it, couldn't we use the same formula but change up those values. In other words, right during deceleration during that phase of deceleration, no, we have our initial velocity. OK. We know that our final velocity is zero. So VF equals 0 m per second. We know our distance. In this case, let's call it H prime or height prime is going to be 1.5 m. And finally, we want to figure out the acceleration so we can use the same formula with these values. OK? Because now using that then that means our final velocity VF squared equals our initial velocity squared. OK plus two multiplied by the acceleration. In this case, the deceleration multiplied by H prime or new distance, we know that we have squared is zero. So that's zero squared equals V not squared plus two A H prime. OK. And no, that means to A H equals negative V not squared, that's H prime. And our acceleration then is going to be equal to negative V not squared divided by two H prime. So now we can find our acceleration that we need to find our force that's being exerted. Because if we use that idea, then that means our acceleration is going to be equal to negative 19.8 m per second squared or sorry, that only the only our velocity, our initial velocity is squared, my bad and all of that is being divided by two multiplied by the height of the snow bunk. In this case, 1.5 m again, let's put this into our calculator. OK? And when you do that, you should get the acceleration to be equal to negative 130.68 m per second squared. Now this is pretty important. What does this mean for us? Well, if we go back to our diagram, no, we know the acceleration during or rather we know the deceleration, we know what the deceleration is during this period. And now we know the deceleration, OK? Sorry, it would be acting against the direction of the climber. Now that we know our deceleration, we can figure out the force exerted by the snow bank. So therefore what this means, let me come over here. OK. Therefore, given that we have the acceleration and we know the climber's mass, that force is going to be equal to his mass. 60 kg multiplied by the deceleration, negative 130.68 m per second squared. And if we figure out if we work out what that force is, put that into our calculator, we should get the force to be equal to negative 7840.8 newtons. Notice all our answers are represented in standard form. So we can write that in standard form as negative 7.8 multiplied by 10 cubed newtons. Now the negative sign indicates the direction of the force is opposite to the direction of the motion. But remember we wanted the magnitude of our decelerating force. So the magnitude of our decelerating force won't have any s and it will just be equal to 7.8 multiplied by 10 cubed newtons. C is the correct answer. Thanks a lot for watching everyone. I hope this video helped

Struggling with Physics?

Suppose a 65-kg person jumps from a height of 3.0 m down to the ground.
(c) Estimate the decelerating force if the person lands stiff-legged so d ≈ 1.0 cm .
<IMAGE>

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Struggling with Physics?

Suppose a 65-kg person jumps from a height of 3.0 m down to the ground.(c) Estimate the decelerating force if the person lands stiff-legged so d ≈ 1.0 cm .<IMAGE>

Watch next

Forces & Kinematics practice set

Suppose a 65-kg person jumps from a height of 3.0 m down to the ground.
(c) Estimate the decelerating force if the person lands stiff-legged so d ≈ 1.0 cm .
<IMAGE>