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.
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].
To find the expression for the crate's speed v when it is at height h above the bottom of the slope, we can use the work-energy principle.
1. The work done on the crate by the horizontal force F is given by W = Fd, where d is the displacement in the direction of the force. Since the crate moves up the slope, the displacement can be expressed as d = h/sin(theta).
2. The work done on the crate against gravity is given by W = mgh, where g is the acceleration due to gravity and h is the height above the bottom of the slope.
3. The total work done on the crate is the sum of the work done by the horizontal force and the work done against gravity. Therefore, we can write:
F * h/sin(theta) + mgh = (1/2)mv^2
4. Rearranging the equation, we get:
F * h/sin(theta) = (1/2)mv^2 - mgh
5. Multiplying both sides of the equation by 2/sin(theta), we obtain:
2Fh = mv^2 - 2mgh
6. Adding 2mgh to both sides of the equation, we have:
mv^2 = 2Fh + 2mgh
7. Finally, dividing both sides of the equation by m, we get the expression for the crate's speed v:
v^2 = (2Fh + 2mgh) / m
So, the expression for the crate's speed v when it is at height h above the bottom of the slope is:
v = sqrt((2Fh + 2mgh) / m)
To answer this question, we can use the principle of work and energy. The work-energy principle states that the net work done on an object is equal to the change in its kinetic energy.
In this case, the only force doing work on the crate is the horizontal force F. Since there is no friction, the only other force acting on the crate is its weight, mg, which is perpendicular to the displacement. Therefore, the work done by the weight of the crate is zero.
The work done by the applied force F is equal to the force multiplied by the distance over which it is applied. The distance the crate moves along the slope can be found using trigonometry. It is equal to h / sin(theta), where h is the vertical height and theta is the angle of the slope.
The work done by the applied force is therefore given by:
Work = F * (h / sin(theta))
According to the work-energy principle, this work done on the crate is equal to the change in its kinetic energy. Initially, the crate is at rest, so its initial kinetic energy is zero. At height h, when it reaches its final speed v, its kinetic energy is (1/2)mv^2.
We can now equate the work done to the change in kinetic energy:
F * (h / sin(theta)) = (1/2)mv^2
Simplifying, we can solve for v by rearranging the equation:
v = sqrt((2Fh) / (m * sin(theta)))
So the expression for the crate's speed v when it is at height h above the bottom of the slope is:
v = sqrt((2Fh) / (m * sin(theta)))