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An 89 kg person climbs up a uniform 13 kg ladder. The ladder is 5.9 m long; its lower end rests on a rough horizontal floor (sta

lapo4ka [179]

Answer:

The maximum distance the person will reach before he slip is 1.82m.

Not even close to the middle of the ladder.

Explanation:

Check attachment for solution

8 0

3 years ago

A boulder is raised above the ground, so that its potential energy relative to the ground is 200 J. Then it is dropped. Estimate

babymother [125]

Answer:

200 J

Explanation:

In this problem, I assume there is no air resistance, so the system is isolated (=no external forces).

For an isolated system, the total mechanical energy is constant, and it is given by:

$E=KE+PE$

where

KE is the kinetic energy

PE is the potential energy

The kinetic energy is the energy due to the motion of the object, while the potential energy is the energy due to the position of the object relative to the ground.

At the beginning, when the boulder is raised above the ground, its height above the ground is maximum, while its speed is zero; it means that all its mechanical energy is just potential energy, and it is:

$E=PE_{max}=200 J$

As the boulder falls down, its altitude decreases, so its potential energy decreases, while the speed increases, and the kinetic energy increases. Therefore, potential energy is converted into kinetic energy.

Eventually, just before the boulder hits the ground, the height of the object is zero, and the speed is maximum; this means that all the energy has now converted into kinetic energy, and we have

$E=KE_{max}=200 J$

Therefore, the kinetic energy just before hitting the ground is 200 J.

6 0

3 years ago

600 kg elephant runs at 5 m/s and jumpts onto a 100kg cart. If the coefficient of friction is .04 how far will the cart travel o

Minchanka [31]

Answer:

Eleven seconds.

Explanation:

Two keys are needed to solve this problem. First, the conservation of momentum: allowing you to calculate the cart's speed after the elephant jumped onto it. It holds that:

$m_ev_e+m_c\cdot0=(m_e+m_c)v_0\implies \\v_0=\frac{m_ev_e}{m_e+m_c}=\frac{600kg\cdot 5\frac{m}{s}}{700kg}=4.29\frac{m}{s}$

So, once loaded with an elephant, the cart was moving with a speed of 4.29m/s.

The second key is the kinematic equation for accelerated motion. There is one force acting on the cart, namely friction. The friction acts in the opposite direction to the horizontal direction of the velocity v0, its magnitude and the corresponding deceleration are:

$F_r = 0.04\cdot (m_e+m_c)g\implies a_r = 0.04\cdot g= 0.04 \cdot 9.8 \frac{m}{s^2}=0.39\frac{m}{s^2}$