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3 years ago

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A 70.0 kg person jumping down from a 3.00 meter wall hits the ground at 7.70 m/s. He hits the ground with stiff legs and stops i

n 0.023 seconds. What is the force of impact, in newtons?

1 answer:

shusha [124]3 years ago

3 0

Answer:

$F_{net} = 24434.783\,N$

Explanation:

Let consider that the person hit the ground in the negative direction. The physical model for the impact is modelled after the Impact Theorem:

$m_{person}\cdot v + Imp = 0$

The impact is:

$Imp = - m_{person}\cdot v$

$Imp = - (70\,kg)\cdot (-7.70\,\frac{m}{s} )$

$Imp = 539\,N\cdot s$

The force of impact is derived as follows:

$F_{net} = \frac{Imp}{\Delta t}$

$F_{net} = \frac{539\,N\cdot s}{0.023\,s}$

$F_{net} = 24434.783\,N$

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Read 2 more answers

A merry-go-round with a rotational inertia of 600 kg m2 and a radius of 3.0 m is initially at rest. A 20 kg boy approaches the m

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Answer:

The velocity of the merry-go-round after the boy hops on the merry-go-round is 1.5 m/s

Explanation:

The rotational inertia of the merry-go-round = 600 kg·m²

The radius of the merry-go-round = 3.0 m

The mass of the boy = 20 kg

The speed with which the boy approaches the merry-go-round = 5.0 m/s

$F_T \cdot r = I \cdot \alpha = m \cdot r^2 \cdot \alpha$

Where;

$F_T$ = The tangential force

I = The rotational inertia

m = The mass

α = The angular acceleration

r = The radius of the merry-go-round

For the merry go round, we have;

$I_m \cdot \alpha_m = I_m \cdot \dfrac{v_m}{r \cdot t}$

$I_m$ = The rotational inertia of the merry-go-round

$\alpha _m$ = The angular acceleration of the merry-go-round

$v _m$ = The linear velocity of the merry-go-round

t = The time of motion

For the boy, we have;

$I_b \cdot \alpha_b = m_b \cdot r^2 \cdot \dfrac{v_b}{r \cdot t}$

Where;

$I_b$ = The rotational inertia of the boy

$\alpha _b$ = The angular acceleration of the boy

$v _b$ = The linear velocity of the boy

t = The time of motion

When the boy jumps on the merry-go-round, we have;

$I_m \cdot \dfrac{v_m}{r \cdot t} = m_b \cdot r^2 \cdot \dfrac{v_b}{r \cdot t}$

Which gives;

$v_m = \dfrac{m_b \cdot r^2 \cdot \dfrac{v_b}{r \cdot t} \cdot r \cdot t}{I_m} = \dfrac{m_b \cdot r^2 \cdot v_b}{I_m}$

From which we have;

$v_m = \dfrac{20 \times 3^2 \times 5}{600} = 1.5$

The velocity of the merry-go-round, $v_m$ , after the boy hops on the merry-go-round = 1.5 m/s.

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