A section of track for a roller coaster consists of two circular arcs AB and CD joined by a straight portion BC. The radius of AB is 27 m and the radius of CD is 72 m. The car and its occupants, of total mass 250 kg, reach point A with practically no velocity and then drop freely along the track. Determine the normal force exerted by the track on the car as the car reaches point B. Ignore air resistance and rolling resistance.

N = 731 N

Explanation:

From the question, we recall the following

The mass m = 250 kg

The initial kinetic energy  T₁ = 0

The initial kinetic energy  T₂ = 1/2 mv²

velocity = v at B

thus,

The work is carried out because of the weight

so,

U₁→₂ = W * h

= (250 * 9.81) * (27 m - 27 cos 40 m)

which is =15491.95 N.m

From the energy principle - from work

T₁ + U₁→₂ = T₂

We can say,

0 + 15491.95 N.m  = 1/2 * 250 kg * v²

v = 11.13 m/s

Now we solve for the normal acceleration

aₙ = v²ρ = (11.13 m/s)²/27 m

aₙ  = 4.588 m/s²

Now, by applying the second law

∑Fy = -maₙ

or we say that,  N - W cos 40° = -maₙ

so,

N = 250 * 9.81 N cos 40° - 250 kg * 4.588 m/s²

Therefore N = 731 N

your answer is, orbit around the nucleus