A 10 kg mass is sitting on a 25 degree incline. A constant force of 49N is applied up the plane. The coefficient of static friction is .25. Does it go up or down or does it stay put?
To determine whether the 10 kg mass will go up, down, or stay put on the 25-degree incline, we need to calculate the net force acting on the mass.
1. Start by finding the gravitational force acting on the mass. The gravitational force is given by the formula Fg = m * g, where m is the mass (10 kg) and g is the acceleration due to gravity (9.8 m/s²).
Fg = 10 kg * 9.8 m/s²
Fg = 98 N
2. Determine the component of the gravitational force acting parallel to the incline. This can be calculated using the formula F_parallel = Fg * sin(θ), where θ is the angle of the incline (25 degrees).
F_parallel = 98 N * sin(25°)
F_parallel ≈ 41.83 N
3. Calculate the force of static friction. The force of static friction can be determined using the equation Fs ≤ μs * N, where μs is the coefficient of static friction and N is the normal force acting perpendicular to the incline. The normal force can be calculated using the formula N = m * g * cos(θ).
N = 10 kg * 9.8 m/s² * cos(25°)
N ≈ 88.87 N
Fs ≤ 0.25 * 88.87 N
Fs ≤ 22.22 N
4. Compare the applied force to the net force. Since the applied force is 49 N and the sum of the force parallel to the incline (41.83 N) and the force of static friction (22.22 N) is less than 49 N, the net force is not enough to overcome the force of static friction. Therefore, the mass will stay put and not move up or down the incline.
In conclusion, the mass will stay put on the 25-degree incline and not move.