A 5.00 kg crate is on a 21.0° hill.
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Answer: The final speed, ignoring the effects of friction, will be 2.88 m/s.
Explanation:
The problem can be solved using two different physical principles. If we resort to Newton's second Law, we can say that, neglecting friction (which is reasonable for a ice block) the only force acting on the block in the direction of the movement along the plane, is the component of the weight that is parallel to the slope, i.e.
Fp = mg sin 28.3º = ma ⇒ a=g sin 28.3º
Now, as we know that g = constant, we can use the following kinematic equation:
vf² - v₀² = 2 a x
if the block starts from rest, this means that v₀ = 0.
Replacing by the values in the equation, and solving for vf, we get;
vf = = 2.88 m/s
The other approach is using the conservation of energy principle:
When the block starts, it has some potential energy = mgh
This height h, can be expressed in terms of x (the length travelled by the block downward the plane) and the angle that forms with the horizontal, as follows:
h = x sin 28.3 (applying sin definition) ⇒ U = mg x sin 28.3
At the the end of the slide, the potential energy has been converted to kinetic energy, so we can write the following equation:
m. g. x. sin 28.3º = 1/2 m vf²
Simplifying, replacing by the values and solving for vf, we arrive to the same result as above.