Doug pushes a 4.65 kg crate up a 2.00 m high 13.0° frictionless slope by pushing it with a constant horizontal force of 35.8 N. What is the speed (in m/s) of the crate as it reaches the top of the slope assuming that the local acceleration due to gravity is -9.80 m/s2?
To solve this problem, we can use the principles of work and energy. The work-energy principle states that the work done on an object is equal to the change in its kinetic energy.
First, let's find the work done by Doug on the crate. The work done is given by the formula:
Work = Force * Distance * Cos(angle)
In this case, the force exerted by Doug is 35.8 N, the distance traveled along the slope is 2.00 m, and the angle of the slope is 13.0°.
Work = 35.8 N * 2.00 m * Cos(13.0°)
Next, let's find the change in potential energy of the crate as it moves up the slope. The potential energy is given by:
Potential Energy = Mass * Gravity * Height
In this case, the mass of the crate is 4.65 kg, the local acceleration due to gravity is -9.80 m/s2, and the height of the slope is 2.00 m.
Potential Energy = 4.65 kg * -9.80 m/s2 * 2.00 m
Since the slope is frictionless, all the work done by Doug is converted into the change in potential energy of the crate.
Work = Change in Potential Energy
Now, let's equate the two equations:
35.8 N * 2.00 m * Cos(13.0°) = 4.65 kg * -9.80 m/s2 * 2.00 m
Simplifying and solving for Work:
Work = 313.32 N·m
Since work is equal to the change in kinetic energy, we can find the final kinetic energy:
Kinetic Energy = Work = 313.32 N·m
Finally, we can use the formula for kinetic energy to solve for the final velocity (speed) of the crate:
Kinetic Energy = (1/2) * Mass * Velocity^2
Simplifying and solving for Velocity:
Velocity = sqrt((2 * Kinetic Energy) / Mass)
Substituting the values:
Velocity = sqrt((2 * 313.32 N·m) / 4.65 kg)
Calculating the result gives:
Velocity ≈ 7.10 m/s
Therefore, the speed of the crate as it reaches the top of the slope is approximately 7.10 m/s.