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2 years ago

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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 stri

ng 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

1 answer:

san4es73 [151]2 years ago

6 0

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

$T - mg cos40 = F_c$

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

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

$T = 3 + 5.55$

$T = 8.55 N$

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