The coefficient of kinetic friction for a 25 -kg bobsled on a track is 0.10.What force is required to push it down a 6.0^\circ incline and achieve a speed of 67 km/h at the end of 75 {\rm m}?
To find the force required to push the bobsled down the incline, we can start by analyzing the forces acting on the bobsled.
1. Gravitational force (Fg): This force is pulling the bobsled down the ramp and can be calculated using the formula Fg = m * g, where m is the mass of the bobsled (25 kg) and g is the acceleration due to gravity (9.8 m/s^2).
Fg = 25 kg * 9.8 m/s^2 = 245 N
2. Normal force (Fn): This force is perpendicular to the incline and acts to support the weight of the bobsled. On an inclined plane, the normal force can be calculated as Fn = m * g * cos(θ), where θ is the angle of the incline (6 degrees).
Fn = 25 kg * 9.8 m/s^2 * cos(6 degrees) = 242.21 N
3. Frictional force (Ff): This force opposes the motion of the bobsled down the incline. The force of friction can be determined using the formula Ff = μ * Fn, where μ is the coefficient of kinetic friction (0.10).
Ff = 0.10 * 242.21 N = 24.22 N
Now, the total force acting on the bobsled down the incline is the sum of the force due to gravity and the force due to friction:
Total force = Fg + Ff = 245 N + 24.22 N = 269.22 N
Therefore, a force of 269.22 Newtons is required to push the bobsled down the incline and achieve a speed of 67 km/h at the end of 75 m.