The work done by the horizontal force F on the crate when the crate moves a distance d along the slope is given by:
W = F * d * cos(theta)
where d is the distance the crate moves along the slope, and theta is the angle between the force F and the direction of displacement of the crate (i.e. parallel to the ramp). Since theta is also the angle of the slope (given the problem), we have:
d * sin(theta) = h (the height to which the crate is lifted)
Solving for d gives us:
d = h / sin(theta)
Substituting d back into the equation for the work done gives us:
W = F * (h / sin(theta)) * cos(theta) = F * h * cos(theta) / sin(theta).
Now, let's look at the change in the kinetic and potential energy of the crate. The work done on the crate is equal to the change in its kinetic and potential energy. The kinetic energy of the crate when it is at height h is given by:
KE = 1/2 * m * v^2.
When the crate is at height h, its potential energy is given by:
PE = m * g * h.
Initially, the crate is at rest, i.e., KE_initial = 0, and PE_initial = 0. Therefore, the change in kinetic and potential energy is:
Delta_KE = KE - KE_initial = 1/2 * m * v^2,
Delta_PE = PE - PE_initial = m * g * h.
So, the work done on the crate can be written as:
W = Delta_KE + Delta_PE = 1/2 * m * v^2 + m * g * h.
We already found that W = F * h * cos(theta) / sin(theta). So, we can write:
F * h * cos(theta) / sin(theta) = 1/2 * m * v^2 + m * g * h.
Now, we can solve for v:
v^2 = 2 * (F * h * cos(theta) / sin(theta) - m * g * h) / m.
Taking the square root of both sides gives:
v = sqrt[2 * (F * h * cos(theta) / sin(theta) - m * g * h) / m].
Starting from rest, a crate of mass m is pushed up a frictionless slope of angle (theta) by a horizontal force of magnitude F. Use work and energy to find an expression for the crate's speed v when it is at height h above the bottom of the slope.
Express your answer in terms of the variables m, F, theta, h, and free fall acceleration g.
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