Amass m= 1.4 kg is hanging from a long string of length l-20.8 m and total mass mi = 13 kg. a student gives the mass a quick horizontal shake to set up a wave which travels upwards along the string 25% part (a) what is the speed of the wave at the bottom of the string, in meters per second? 25% part (b) express the tension of the chain at a height y above the bottom in terms of m, m, and g 25% part (c) what is the speed of the wave, in meters per second, at the top of the string?

a. v = 4.68 m/s²

b. T₂ = 141.12 N

c. v = 15.02 m/s²

Explanation:

Given:

M= 1.4 kg, L = 20.8m, mi=13 kg

a.

Using the equations of kinetic energy and the spring potential

Uk = Ek

¹/₂ * T * d =  ¹/₂ * m * v ²

T₁ = M * g  = 1.4 kg * 9.8 m/s² = 13.72 N

T₂ =  (M + mi) * g = (1.4 kg + 13 kg)* 9.8 m/s² = 141.12 N

a. the speed at the bottom

v = √[(T * d )/ m] = √ 13.72 N * 20.8 m / 13 kg  

v = 4.68 m/s²

b. the tension of the chain

T₂ =  (M + mi) * g = (1.4 kg + 13 kg)* 9.8 m/s² = 141.12 N

c. the speed at the top of the spring

v = √[(T * d )/ m] = √ 141.12 N * 20.8 m / 13 kg  

v = 15.02 m/s²

the box does not move because you need to use a force greater than 300 n. you are only using 50 n. it needs to at least be 301 n for the box to move because then it is an unbalanced force and it will move the way that the greater force is pushing it. if you use only 300 n it will be a balanced force and will not move because the force you are exerting on the   box is the same as its force.

hope this !

im sorry if i went too in depth, i am going to include a link to a video on khan academy that will explain this.   hope that you like the video and that it explains more in detail to you.

that is the link.

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