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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

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.

How do I solve for v

Quote:

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.