A 5.05kg sits on a ramp that is inclined at 36.5 degrees above horizontal.The coefficient of kinetic friction between box and ramp is mu k= 0.25. What horizontal force is required to move the box up the incline with a constsant acceleration of 4.3m/s^2?
The NET force up the incline must be
Fnet = M*a = 5.05*4.3 = 21.72 N
This equals the sum of of the applied force component up the ramp (Fap*cos36.5), minus the friction force (M*g*muk*cos36.5-Fap*sin36.5muk)) and minus the weight component down the ramp (M*g*sin36.5).
Solve for Fap
To determine the horizontal force required to move the box up the incline with a constant acceleration, we need to follow these steps:
Step 1: Resolve the forces.
- First, we need to resolve the weight of the box into its components.
- The weight of the box can be calculated using the formula: weight = mass * gravitational acceleration.
- In this case, weight = 5.05 kg * 9.8 m/s^2 ≈ 49.49 N.
- The weight can be divided into two components: the component parallel to the incline (mg*sinθ) and the component perpendicular to the incline (mg*cosθ).
- For the given angle of 36.5 degrees, the parallel component (mg*sinθ) = 49.49 N * sin(36.5) ≈ 29.37 N, and the perpendicular component (mg*cosθ) = 49.49 N * cos(36.5) ≈ 39.68 N.
Step 2: Calculate the frictional force.
- The coefficient of kinetic friction (μk) is given as 0.25.
- The frictional force (f) can be calculated using the formula: f = μk * (mg*cosθ).
- Therefore, f = 0.25 * 39.68 N ≈ 9.92 N.
Step 3: Apply Newton's second law.
- The net force acting on the box is the difference between the applied force (F) and the frictional force (f). So, net force (F_net) = F - f.
- According to Newton's second law, F_net = mass * acceleration.
- Substituting the given values, F - 9.92 N = 5.05 kg * 4.3 m/s^2.
Step 4: Calculate the horizontal force.
- Rearrange the equation to solve for F: F - 9.92 N = 5.05 kg * 4.3 m/s^2.
- F = (5.05 kg * 4.3 m/s^2) + 9.92 N.
- F ≈ 26.22 N.
Therefore, the horizontal force required to move the box up the incline with a constant acceleration of 4.3 m/s^2 is approximately 26.22 N.