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bayley86 · Dec 9, 2010, 12:15 PM

A box weighing 2500kg slides from a slope of length 10 m and height of 2 m. determine the velocity of the box when it reaches the bottom

ebaines · Dec 9, 2010, 03:14 PM

Use conservation of energy. The gain in kinetic energy of the box when it reaches the bottom must equal the loss of potential energy from sliding downhill. If h is the height of the ramp, and assuming the ramp is frictionless and the box starts with velocity =0:

1/2mv^2 = mgh

Solve for v.

bayley86 · Dec 14, 2010, 10:34 AM

How do I solve for v

ebaines · Dec 14, 2010, 12:07 PM

Originally Posted by bayley86

how do i solve for v

Basic algebraic manipulation:

Starting with the equation I gave you earlier:

$ \frac 1 2 m v^2 = mgh $

Divide through by m:
$ \frac 1 2 \cancel{m} v^2 =\cancel {m}gh \\ \frac 1 2 v^2 = gh $

multiply through by 2:

$ 2 \times (\frac 1 2 v^2) = 2 \times gh\\ v^2 = 2 gh $

Take the square root of both sides:

$ \sqrt{v^2} = \sqrt{2 gh} \\ v = \sqrt {2gh} $

You've been given the value of h, which is 2 meters. And g is acceleration due to gravity, which is 9.8 m/s^2.

Hope this helped.