A small rock with mass 0.12 kg is fastened to a massless string with length 0.80 m to form a pendulum. The pendulum is swinging so as to make a maximum angle of 45 ∘ with the vertical. Air resistance is negligible.

A. What is the speed of the rock when the string passes through the vertical position?

Express your answer using two significant figures.

B. What is the tension in the string when it makes an angle of

45∘ with the vertical?

Express your answer using two significant figures.

C. What is the tension in the string as it passes through the vertical?

Express your answer using two significant figures.

The answers to the question are

(a) 2.1 m/s

(b) 0.83 N

(c) 1.9 N

Explanation:

To solve the question, we list out the varibles

Length, l of string = 0.8 m

mass of rock, m = 0.12 kg

Angle with the verrticakl, θ = 45 °

a) To find the speed of the rock  when the string passes through the vertical position we have

From the first law of thermodynamics

Potential energy = kinetic energy

m×g×l×(1-cosθ) = 1/2×m×v²

That is v² = 2×g×l×(1-cosθ)

= 2×9.81×0.8×(1-cos45) = 4.597

or v = √4.597 = 2.1 m/s

(b) The tension in the string when it makes an angle of  45∘ with the vertical is given by

For balance between Tension and mass of rock is gigen by

∑Forces = 0, T - m×g×cosθ = 0

or T =  m×g×cosθ = 0.12×9.81×cos45 = 0.83 N

c) The tension in the string as it passes through the vertical

when passing through the vertical we have T - m×g = (m×v²)/r

or T = m×g + (m×v²)/r = mg(1+2(1-cosθ)) =0.981*0.12 (1+ 2(1-cos45)) =1.867 N

= 1.9 N

explanation:

v=uf/(u-f)

= 5.-10/(5+10) convex lens has negative focal length