Question

A 0.60-kg object is suspended from the ceiling at the end of a 2.0-m string. When pulled to the side and released, it has a speed of 4.0 m/s at the lowest point of its path. What maximum angle does the string make with the vertical as the object swings up?

Answer 1

Answer:

ø = 53.7º

Explanation:

The measurement of kinetic energy in an object is calculated based on the object's mass and velocity.

KE=1/2mv²

where m is mass

v is velocity

given data

m=0.60 kg

v=4.0 m/s

So

KE = 1/2 . 0.6 . 4^2  

KE= 4.8 J

All the KE is converted into GPE ("Gravitational potential energy (GPE) - energy stored in an object when moving the object to a height")

4.8  = 0.6 . 9.8 . ∆h

∆h  = 0.816 m

This is the height that the object rises.

cos ø  = (2 – 0.816 ) / 2 = 0.592

ø = 53.7º

Answer 2

The angle of the string from the verticle axis is 53.48 degrees.

How do you calculate the angle of the string?

Given that the mass m of the object is 0.60 kg and length l of the string is 2 m. The velocity v of the object is 4 m/s.

The kinetic energy of the object is given below.

$KE = \dfrac {1}{2}mv^2$

$KE = \dfrac {1}{2}\times 0.60\times 4^2$

$KE = 4.8 \;\rm J$

This kinetic energy of the object is converted into gravitational potential energy which is stored in while moving the object to a height.

$GPE = mg\Delta h$

$4.8 = 0.60 \times 9.8 \times \Delta h$

$\Delta h = 0.81$

The angle with the vertical axis is calculated as given below.

$cos \theta = \dfrac {l-\Delta h} {l}$

$cos \theta = \dfrac {2-0.81}{2}$

$cos \theta = 0.595$

$\theta = 53.48^\circ$

Hence we can conclude that the angle of the string from the verticle axis is 53.48 degrees.

To know more about kinetic and potential energy, follow the link given below.

https://brainly.com/question/21288807.