Verified answer

Let L be the length of the rope. When the basket is pulled back 50°, the height of the basket is increased by an amount L - L cos 50°. Taking the zero of gravitational potential energy to be at the lowest point of the rope, we can compute that, when a person of mass M sits in the basket, the potential energy of the occupied basket is:

U = (M + 5kg)g(L - L cos 50°)

= (M+5)(1 - cos 50°)gL

At the lowest part of the swing, U = 0, and since gravity is conservative, the kinetic energy must equal (M+5)(1 - cos 50°)gL:

(1/2)(M + 5)v² = (M + 5)(1 - cos 50°)gL

v² = 2gL(1 - cos 50°)

Consider a free-body diagram when the basket is at the lowest part of the swing. Two forces act on the basket: the gravitational force of magnitude (M + 5)g acting downward, and the tension T in the rope acting upward. The sum of these two forces must be equal to the centripetal force causing the basket to swing in circular motion:

T - (M + 5)g = (M + 5)v²/L

T = (M+5)(2gL(1 - cos 50°))/L + (M + 5)g

= (M+5)g[(2 - 2 cos 50°) + 1]

= (M + 5)g(3 - 2 cos 50°)

We require this tension to be less than 1000 N:

T < 1000

(M+5)(3 - 2 cos 50°)g < 1000

M + 5 < 1000/[(3 - 2 cos 50°)g]

M < 1000/[(3 - 2 cos 50°)g] - 5

M < 54.5 kg