4. A crate is pulled up a ramp a distance of 20 m with a force of F = [800, 500]. The crate ends up 2 m higher above the ground than where it started. a. (APP - 4 marks) Find the force applied on the crate that moves the crate up the ramp and the force that is perpendicular to the ramp that lifts the crate up off the ramp. b. (APP - 2 marks) Find the work done moving the crate up the ramp and the work done lifting the crate 2 m above the ground.
Question
- A crate is pulled up a ramp a distance of 20 m with a force of F = [800, 500]. The crate ends up 2 m higher above the ground than where it started.
a. (APP - 4 marks) Find the force applied on the crate that moves the crate up the ramp and the force that is perpendicular to the ramp that lifts the crate up off the ramp.
b. (APP - 2 marks) Find the work done moving the crate up the ramp and the work done lifting the crate 2 m above the ground.
Solution
a. The force applied on the crate that moves the crate up the ramp and the force that is perpendicular to the ramp that lifts the crate up off the ramp can be found by decomposing the force vector F into its components parallel and perpendicular to the ramp.
First, we need to find the angle of the ramp. We can use the Pythagorean theorem to find the hypotenuse of the triangle formed by the ramp, the height, and the horizontal distance. The hypotenuse is sqrt(20^2 + 2^2) = sqrt(404) = 20.1 m.
The angle of the ramp is then sin^-1(2/20.1) = 5.7 degrees.
The component of the force F parallel to the ramp is F_parallel = F * cos(theta) = sqrt(800^2 + 500^2) * cos(5.7) = 938.5 N.
The component of the force F perpendicular to the ramp is F_perpendicular = F * sin(theta) = sqrt(800^2 + 500^2) * sin(5.7) = 93.8 N.
b. The work done moving the crate up the ramp is W_ramp = F_parallel * distance = 938.5 N * 20 m = 18770 J.
The work done lifting the crate 2 m above the ground is W_lift = F_perpendicular * height = 93.8 N * 2 m = 187.6 J.
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