Question

A 30 g toy train is pushed up against a horizontal spring and compresses the spring by a distance of 13cm and then is released. The train reaches an incline of 27.2 degrees that has a coefficient of kinetic friction of 0.4. How far up the ramp will the car go?

Answer 1

here,

mass , m = 30 g = 0.03 kg

x = 13 cm = 0.13 m

let the inital speed be u

using conservation of energy

0.5 * K * x^2 = 0.5 * m * u^2

0.5 * 62.2 * 0.13^2 = 0.5 * 0.03 * u^2

solving for u

u = 5.92 m/s

theta = 27.2 degree

uk = 0.4

accelration , a = - ( g * sin(theta) + uk * g * cos(theta))

a = - ( 9.81 * sin(27.2) + 0.4 * 9.81 * cos(27.2))

a = - 7.97 m/s^2

let the distance up the incline be s

using third equation of motion

v^2 - u^2 = 2 * a * s

0 - 5.92^2 = - 2 * 7.97 * s

solving for s

s = 2.2 m

the distance up the incline is 2.2 m