A 50kg block is being pulled up a 13 degree slope by a force directed 30 degrees above the slope. The coeffcienct of kinetic friction between the block and the slope is 0.200. What is the minimum force required for the block to continue to slide up the plane with no acceleration once it has been set in motion?
To find the minimum force required for the block to continue to slide up the slope with no acceleration, we need to consider the different forces acting on the block.
Let's break down the forces:
1. Weight force (W): This force acts vertically downwards and is given by the equation W = mg, where m is the mass of the block and g is the acceleration due to gravity. In this case, m = 50 kg.
W = 50 kg * 9.8 m/s^2 = 490 N
2. Normal force (N): This force acts perpendicular to the slope and prevents the block from sinking into the slope. It can be calculated using the equation N = mg*cos(θ), where θ is the angle of the slope. In this case, θ = 13 degrees.
N = 50 kg * 9.8 m/s^2 * cos(13 degrees) = 474.99 N
3. Force of friction (f): This force acts parallel to the slope and opposes the block's motion. It can be calculated using the equation f = μ*N, where μ is the coefficient of kinetic friction. In this case, μ = 0.200.
f = 0.200 * 474.99 N = 94.999 N
4. Force applied (F): This is the force we want to find. It acts at an angle of 30 degrees above the slope.
Now we can analyze the forces in the x and y directions separately.
In the y-direction (perpendicular to the slope), the sum of forces is given by:
N - W*cos(30 degrees) = 0
474.99 N - 490 N * cos(30 degrees) = 0
Simplifying the equation above, we can find the value of N.
In the x-direction (parallel to the slope), the sum of forces is given by:
F - W*sin(30 degrees) - f = 0
F - 490 N * sin(30 degrees) - 94.999 N = 0
Simplifying the equation above, we can find the value of F, which is the minimum force required for the block to continue sliding up the slope with no acceleration.