A 43.0-g toy car is released from rest on a frictionless track with a vertical loop of radius R. The initial height of the car is h = 4.05R. Required:
a. What is the speed of the car at the top of the vertical loop?
b. What is the magnitude of the normal force acting on the car at the top of the vertical loop?

A.) 909 cm/s

B.) 33075 N

Explanation:

A.) Given that the

Mass M = 43 g

Height h = 4.05 R

Radius r = R

At the top of the loop, the maximum potential energy P.E = mgh

Substitutes m and h into the formula where g = 9.8 m/s^2 = 9610.517 cm/s^2

P.E = 43 × 9610.517 × 4.05R

P.E = 1673671.536R J

According to conservative of energy

The maximum P.E = maximum K.E

But K.E = 1/2mv^2

1673671.536R = 1/2mv^2

Substitutes for mass m into the formula

1673671.536R = 1/2× 4.05R × v^2

The R will cancel out

Cross multiply

4.05 v^2 = 3347343.072

V^2 = 3347343.072 / 4.05

V^2 = 826504.4622

V = sqrt( 826504.4622)

V = 909 cm/s

B.) At the top of the loop, the centripetal force = the sum of the normal force N and the weight W of the car. That is,

MV^2/R = N + W

Make N the subject of formula

N = mv^2/ R - W

Where W = mg

Substitute all the parameters into the formula

N = (4.05R × 909^2) /R - 4.05R × 9610.517

N = 3346438.05 - 38922.59

N = 3307515 N

energy stored in the capacitor= 9.72 x 10⁻⁴ j

explanation:

the energy stored in the capacitor is given by

e= 1/2 c v²

e= energy stored

c= capacitance= 1.5 µc= 1.5 x 10⁻⁶f

v= potential difference=36 v

so energy stored= e= 1/2 (1.5 x 10⁻⁶) (36)²

energy stored= 9.72 x 10⁻⁴j