Question

A 0.40-kg mass attached to the end of a string swings in a vertical circle having a radius of 1.8 m. At an instant when the string makes an angle of 40 degrees below the horizontal, the speed of the mass is 5.0 m/s. What is the magnitude of the tension in the string at this instant?
a. 9.5 N
b. 3.0 N
c. 8.1 N
d. 5.6 N
e. 4.7 N

Answer 1

Answer:

T = 8.55 N

Explanation:

When string makes an angle 40 degree with the vertical then it will have two forces on it

1) gravitational force (mg)

2) Tension force in string (T)

now we know that net force towards the center of the path is known as centripetal force and it is given as

$T - mg cos40 = F_c$

$T - (0.40\times 9.8)cos40 = \frac{mv^2}{L}$

$T = 3 + \frac{0.40\times 5^2}{1.8}$

$T = 3 + 5.55$

$T = 8.55 N$

Answer 2

A 0.40-kg mass attached to the end of string swings in a vertical circle having a radius of 1.8 m,  the magnitude of the tension in the string at this instant is

T= 8.1 N Option C. This is further explained below.

What is the magnitude of the tension in the string at this instant?

Generally, the tension equation is

$T - mg cos40 = F_c$

Where

$F_c=mv^2/L$

$F_c=(0.40*v^2)/L\\\\F_c=(0.40*5^2)/L\\\\F_c=(0.40*5^2)/1.8\\\\F_c= 5.55........equ1$

Therefore

$T - mg cos40 = F_c$

$T-(0.4*9.81) cos40 = F_c\\\\T-(0.4*9.81) (0.7660444431) = F_c$

When (0.4*9.81) * (0.7660444431) we have

3.005958395

Hence

T =3.005958395  +(0.40*v^2)/L

From equ 1

T = 3.005958395  +5.55

T = 8.1 N

In conclusion, the magnitude of the tension

T = 8.1 N

Read more about tension

https://brainly.com/question/13397436

#SPJ5