Question

A 0.50-kg mass attached to the end of a string swings in a vertical circle (radius 2.0 m). When the mass is at the highest point on the circle, the speed of the mass is 12 m/s. What is the magnitude of the force of the string on the mass at this position?

Answer 1

Answer: 41N

Explanation :

T= mv^2/R + mgcos θ

At the highest point on the circle θ=0

Cos 0 = 1

T= mv^2/R + mg

m = 0.5kg

Velocity at the highest point (amplitude)= 12m/s

T = 0.5× 12^2/2 + 0.5×10

0.5×144/2 +5

T = 0.5×72 + 5

T = 36+5

T = 41N

Answer 2

31.1 N

Step-by-step explanation:

m = mass attached to string = 0.50 kg

r = radius of the vertical circle = 2.0 m

v = speed of the mass at the highest point = 12 m/s

T = force of the string on the mass attached.

At the highest point, force equation is given as

`T + mg =(mv^(2))/(r)`

Inserting the values

`T + (0.50)(9.8) =((0.50)(12)^(2))/(2)`

T = 31.1 N