Problem 19 Medium Difficulty

(III) A person jumps from the roof of a house 2.8 m high. When he strikes the ground below, he bends his knees so that his torso decelerates over an approximate distance of 0.70 m. If the mass of his torso (excluding legs) is 42 kg, find ($a$) his velocity just before his feet strike the ground, and ($b$) the average force exerted on his torso by his legs during deceleration.

Answer

a) 7.4 $\mathrm{m} / \mathrm{s}$
b) $2100 \mathrm{N}$

May 13, 2020

why did you say mg-Ft and not the other way around?

Top Physics 101 Mechanics Educators

Christina K.

Rutgers, The State University of New Jersey

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Video Transcript

In this problem on the topic of dynamics, we are told that a person jumps off the roof of a house that is two m high. When they strike the ground, they bend their knees so that the torso decelerates over a distance of about 0.7 m. The mass of this person's torso is 42 kgs. And we want to use this information to find his velocity just before his feet strike the ground and the average force that his legs will exert on his torso. During this deceleration. Now, we'll use a kind of medical equation of motion to calculate the speed just before he strikes the ground. We know that These initial speed is zero since he starts from rest, the distance that he travels. Why? Minus? Why not is to 0.8 m? And the acceleration is that due to gravity 9.8 m/km2. So we use the equation V squared minus, Vinod squared Is equal to two a. Into why minus? Why not? And rearranging we get the final speed. V. To be the square root of two A into y minus. Why not? Since we know it is equal to zero. Okay, this is the square root Of two times the acceleration due to gravity 9.8 m/km2 Times a distance of two 0.8 m, which gives us his final speed As he approaches the ground to be 7.408 m/s, Which is 7.4 m/s. 2, 2 significant figures. Now for part B who went to find the average force exerted on his torso by his legs during this deceleration. And for this we need the acceleration or the deceleration when coming to rest. So to find the average declaration will choose down to be positive. We have an initial speed, we not Of 7.408 m/s. A final speed when it comes to rest, we is able to zero and the distance over which is also comes to rest. Why minus? Why not? is 0.7 meters? So use the same equation in that V squared -3. Not squared is equal to to a into data. Why? Which means that his deceleration A is minus the north squared over two delta Y. This is -7 For 08 m/s squared, divided by two times 0.7 m. And so calculating, we get his average deceleration two B -39.2 m/km2. Now the average force on the torso will call ft due to the legs is found from Newton's second law. So the net force is equal to his weight downwards. My Minister force on his torso which is upward, and by Newton's second Law, this must be M. Times A. And so the false honest. Also, ft must equal to MG minus M. A. Or simply M. Is mass into G minus eight. And so this is The mass of the torso, 42 kg into 9.8 meters per square second -39.2 m/km2. And so we get the force exerted on the torso during deceleration To be 2100 newtons, and has forced this force act upwards.

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