Question

A .200kg ball is attached to a light 1.2 m long string. The ball is the whirled in a horizontal circle with a constant speed of 1.5m/s. The string makes the angle of 35 degrees with respect to the vertical. What is the tension of the string?

Answer 1

the tension in the string is approximately 0.307 N

Step-by-step explanation:

The tension of the string can be calculated using the centripetal force formula:

F = (m * v^2) / r

where F is the centripetal force, m is the mass of the ball, v is its velocity, and r is the radius of the circular path. In this case, the radius is equal to the length of the string, which is 1.2 m.

First, we need to find the horizontal component of the tension force, which is equal to the centripetal force. We can find this by multiplying the total force by the cosine of the angle between the string and the vertical:

F_horizontal = F * cos(35°)

Next, we can substitute the given values into the centripetal force formula:

F = (m * v^2) / r

F = (0.2 kg * (1.5 m/s)^2) / 1.2 m

F = 0.375 N

Finally, we can substitute this value into the equation for the horizontal component of the tension force:

F_horizontal = F * cos(35°)

F_horizontal = 0.375 N * cos(35°)

F_horizontal = 0.307 N (rounded to three significant figures)