Question

A 0.62 kg car slides 0.24 m down a frictionless ramp at an angle of 30 degrees. It travels across a frictionless surface and into a spring with k = 55 N/m.

Determine the maximum acceleration of the toy car after it hits the spring.

Answer 1

v=sqrt(2gh)

v is the final velocity on hitting the spring

sqrt denotes square root

g is the acceleration due to gravity

h is the height from which the car was released.

We can use this equation as the car is frictionless

h=Sin(30 degrees) x 0.24 = 0.12 m

v=1.53 m/s

this kinetic energy is converted to potential energy in spring

0.5 m v2= 0.5 k x2 ( we can even work without finding v, use mgh=0.5 k x2 )

0.62 x (1.53 x 1.53) = 55 x2

x=0.16m

Now

F=-kx

so

m a = -55 x 0.16

a=-55x0.16/0.62

a=-14.19 m/s2