WAEC - Physics (2024 - No. 11)
A stone of mass 200 g attached to a string is made to revolve in a horizontal circle of radius 1.5 m at a steady speed of 5 m/s. Calculate the tension in the string at the bottom of the circle.
5.3 N
3.3 N
2.0 N
1.3 N
Explanation
F\(_c\) = \(\frac{mv^2}{r}\)
mass = 200g = 0.2kg, v = 5 m/s, r = 1.5 m
F\(_c\) = \(\frac{0.2 \times 5 \times 5}{1.5}\) ≈ 3.33N
At the bottom of the circle, the tension T in the string must counteract the weight of the stone as well as provide the centripetal force:
T = F\(_c\) + mg
T = 3.33 + 0.2 x 10 = 5.3N
Comments (0)
