Question

A 2kg box is at the top of a frictionless ramp at an angle of 60o. The top of the ramp is 30m above the ground. The box is sitting still while at the top of the ramp, and is then released. What is the velocity of the box just before it hits the ground?

Answer 1

Answer:

Explanation:

Given A 2 kg box is at the top of a friction less ramp at an angle of 60 degree. The top of the ramp is 30 m above the ground. The box is sitting still while at the top of the ramp, and is then released. What is the velocity of the box just before it hits the ground?

We need to know the time the box takes to reach the ground. Given vertical distance is 30 m and also angle of the ramp will be

1/2 = 30 m / x

Or x = 60 m

Now the acceleration of the box due to gravity will be  

g sin θ

= 10 sin 60

= 10 x √3 / 2

= 8.66 m / s^2

Now we know that distance equation will be x = xo – 1/2 (g sin θ) t^2

                                                                    0 = 60 – 1/2  (8.66) t^2

                                                                    0 = 120 – 8.66 t^2

                                                                          8.66 t^2 = 120

                                                                          t ^2 = 120 / 8.66

                                                                        t^2 = 13.856

                                                               So t = 3.72 secs

Now we need to use the equation v = u + at

                              So v = 0 + (8.66)(3.72)

                                  So v = 32.2 m/s