A rope pulls a 63.0 kg snow skier up a 14.0 degree incline with a tension of 512 N parallel to the ramp. The coefficient of kinetic friction between the skier and the ramp is 0.270. Calculate the net force and the acceleration of the skier during the moment.
The correct answer and explanation is:
To calculate the net force and acceleration of the skier, we first need to determine the forces acting on the skier along the incline.
1. Forces Acting on the Skier:
The forces involved in this situation are:
- Tension (T): The force pulling the skier up the ramp, which is given as 512 N.
- Gravitational Force (Fg): The component of the skier’s weight that acts down the ramp, given by Fg=mgsin(θ)F_g = mg \sin(\theta), where m=63.0 kgm = 63.0 \, \text{kg} and θ=14.0∘\theta = 14.0^\circ. Fg=63.0 kg×9.8 m/s2×sin(14∘)=63.0×9.8×0.2419=148.6 NF_g = 63.0 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times \sin(14^\circ) = 63.0 \times 9.8 \times 0.2419 = 148.6 \, \text{N}
- Frictional Force (F_f): The force opposing the skier’s motion due to kinetic friction, given by Ff=μk×NF_f = \mu_k \times N, where μk=0.270\mu_k = 0.270 and NN is the normal force. The normal force on an inclined plane is given by N=mgcos(θ)N = mg \cos(\theta). N=63.0 kg×9.8 m/s2×cos(14∘)=63.0×9.8×0.9703=601.6 NN = 63.0 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times \cos(14^\circ) = 63.0 \times 9.8 \times 0.9703 = 601.6 \, \text{N} The frictional force is: Ff=0.270×601.6 N=162.4 NF_f = 0.270 \times 601.6 \, \text{N} = 162.4 \, \text{N}
2. Net Force:
The net force acting on the skier along the incline can be calculated using Newton’s second law. The net force is the difference between the tension in the rope, the gravitational force, and the frictional force: Fnet=T−Fg−FfF_{\text{net}} = T – F_g – F_f Fnet=512 N−148.6 N−162.4 N=201.0 NF_{\text{net}} = 512 \, \text{N} – 148.6 \, \text{N} – 162.4 \, \text{N} = 201.0 \, \text{N}
3. Acceleration:
To calculate the acceleration of the skier, we use Newton’s second law: a=Fnetma = \frac{F_{\text{net}}}{m} a=201.0 N63.0 kg=3.19 m/s2a = \frac{201.0 \, \text{N}}{63.0 \, \text{kg}} = 3.19 \, \text{m/s}^2
Conclusion:
The net force acting on the skier is 201.0 N, and the skier’s acceleration is 3.19 m/s23.19 \, \text{m/s}^2. This means the skier is accelerating up the ramp at this rate under the given conditions.