A crate is given an initial speed of 2.9 up the 29 plane shown in the figure . Assume = 0.13
how far up the plane will it go?
Symbols and units must be provided.
A statement like
"Assume = 0.13"
is meaningless by itself.
Damon has managed to figure out what you probably mean.
LOL except I left out the 2 from the 1/2
To determine how far up the plane the crate will go, we need to calculate the vertical distance covered by the crate.
First, we need to find the component of the initial speed that is directed up the plane. This can be done using the trigonometric relationship between the angle of the plane and the inclined plane:
sin(angle) = opposite/hypotenuse
In this case, the opposite side represents the vertical distance covered by the crate (the distance up the plane), and the hypotenuse represents the initial speed of the crate. Therefore, we can rearrange the equation as follows:
opposite = sin(angle) * hypotenuse
Given that the initial speed is 2.9, and the angle of the plane is 29 degrees, we can plug in these values into the equation:
opposite = sin(29) * 2.9
Using a scientific calculator or a trigonometric table, we can find the sine of 29 degrees, which is approximately 0.4848. Plugging this value into the equation:
opposite = 0.4848 * 2.9 = 1.40472
Therefore, the crate will go approximately 1.40472 units up the plane.
Normal force = m g cos 29
friction force = .13 m g cos 29
call distance along surface headed up = d
work done against friction = .13 m g d cos 29
work done against gravity = m g d sin 29
so
(1/2) m v^2 = .13 m g d cos 29 + m g d sin 29
v^2 =2.9^2 = d (.13 cos 29 + sin 29)
solve for d which is distance sliding up ramp
if you want the final height it is d sin 29