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Klio2033 [76]
3 years ago
10

Chris and Sue are wearing harnesses and are hanging from the ceiling by means of ropes attached to them. They are face to face a

nd push off against each other. Sue has a mass of 52 kg and Chris has a mass of 78 kg. Following the push, Sue reaches a height of 0.65 meters above her starting point. How high does Chris reach?
Physics
1 answer:
Setler [38]3 years ago
3 0

Answer:

h = 0.288m

Explanation:

Assume

v_1 = Speed of Sue

v_2 = Speed of Chris immediately after the push

Sue's KE = \frac{1}{2} mv_1\ ^2 = 26 v_1 \ ^2

now she swings this is converted into gravitation at PE of

mg n = 52 × 9.8 × 0.65  

= 331.24

26v_{1}\ ^2 = 331.24

So, v_1 = 3.569

They started at rest by conservation of momentum in case of push off the magnitude of sue momentum and it is equal to the magnitude of Chris momentum in the opposite or inverse direction

m_1v_1 = m_2v_2

52 \times 3.569  = 78 \times v_2

v_2 = 2.380

Chris kt = \frac{1}{2} \times 78 \times 2.380^2

= 220.827

220.827 = mgh

So, h = 0.288m

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a) The speed of the block at a height of 0.25 m is 2.38 m/s

b) The compression of the spring is 0.25 m

c) The final height of the block is 0.54 m

Explanation:

a)

We can solve the problem by using the law of conservation of energy. In fact, the total mechanical energy (sum of kinetic+gravitational potential energy) must be conserved in absence of friction. So we can write:

U_i +K_i = U_f + K_f

where

U_i is the initial potential energy, at the top

K_i is the initial kinetic energy, at the top

U_f is the final potential energy, at halfway

K_f is the final kinetic energy, at halfway

The equation can be rewritten as

mgh_i + \frac{1}{2}mu^2 = mgh_f + \frac{1}{2}mv^2

where:

m = 2.7 kg is the mass of the block

g=9.8 m/s^2 is the acceleration of gravity

h_i = 0.54 is the initial height

u = 0 is the initial speed

h_f = 0.25 m is the final height of the block

v is the final speed when the block is at a height of 0.25 m

Solving for v,

v=\sqrt{u^2+2g(h_i-h_f)}=\sqrt{0+2(9.8)(0.54-0.25)}=2.38 m/s

b)

The total mechanical energy of the block can be calculated from the initial conditions, and it is

E=K_i + U_i = 0 + mgh_i = (2.7)(9.8)(0.54)=14.3 J

At the bottom of the ramp, the gravitational potential energy has become zero (because the final heigth is zero), and all the energy has been converted into kinetic energy. However, then the block compresses the spring, and the maximum compression of the spring occurs when the block stops: at that moment, all the energy of the block has been converted into elastic potential energy of the spring. So we can write

E=E_e = \frac{1}{2}kx^2

where

k = 453 N/m is the spring constant

x is the compression of the spring

And solving for x, we find

x=\sqrt{\frac{2E}{k}}=\sqrt{\frac{2(14.3)}{453}}=0.25 m

c)

If there is no friction acting on the block, we can apply again the law of conservation of energy. This time, the initial energy is the elastic potential energy stored in the spring:

E=E_e = 14.3 J

while the final energy is the energy at the point of maximum height, where all the energy has been converted into gravitational potetial energy:

E=U_f = mg h_f

where h_f is the maximum height reached. Solving for this quantity, we find

h_f = \frac{E}{mg}=\frac{14.3}{(2.7)(9.8)}=0.54 m

which is the initial height: this is correct, because the total mechanical energy is conserved, so the block must return to its initial position.

Learn more about kinetic and potential energy:

brainly.com/question/1198647

brainly.com/question/10770261

brainly.com/question/6536722

#LearnwithBrainly

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