a 63-kg water skier is pulled up a 14.0 degree incline by a ropeparallel to the incline with a tension of 512 N. the coefficient ofkinetic friction is 0.27. What are the magnitude and direction ofthe skier's acceleration
3.17m/s^2
To find the skier's acceleration, we need to break down the forces acting on the skier in the direction parallel to the incline.
1. Gravitational Force (Weight):
The weight of the skier can be calculated using the formula: Weight = mass × acceleration due to gravity.
Weight = 63 kg × 9.8 m/s² ≈ 617.4 N
2. Normal Force:
The normal force acts perpendicular to the incline and balances the component of the skier's weight perpendicular to the incline. On an inclined plane, the normal force can be calculated using the formula: Normal force = weight × cos(θ), where θ is the angle of the incline.
Normal force = 617.4 N × cos(14°) ≈ 601.7 N
3. Frictional Force:
The frictional force opposes the motion of the skier and can be calculated as the product of the coefficient of kinetic friction and the normal force.
Frictional force = coefficient of kinetic friction × normal force
Frictional force = 0.27 × 601.7 N ≈ 162.5 N
4. Tension Force:
The tension force in the rope is parallel to the incline and can be considered as one of the forces acting on the skier. Its magnitude is given as 512 N.
Now, let's calculate the net force acting on the skier:
Net force = tension force - frictional force
Net force = 512 N - 162.5 N = 349.5 N
Finally, the acceleration of the skier can be calculated using Newton's second law: Force = mass × acceleration.
Acceleration = Net force / mass
Acceleration = 349.5 N / 63 kg ≈ 5.55 m/s²
Therefore, the magnitude of the skier's acceleration is approximately 5.55 m/s². The direction of acceleration is parallel to the incline, which means it is uphill in this situation.
To find the magnitude and direction of the skier's acceleration, we need to consider the forces acting on the skier.
1. Force of gravity (Weight):
The weight of the skier is given by the formula: W = m * g, where m is the mass and g is the acceleration due to gravity (9.8 m/s²).
W = 63 kg * 9.8 m/s² = 617.4 N (downward)
2. Normal force:
The normal force is the force exerted by the incline perpendicular to its surface. It is equal in magnitude and opposite in direction to the component of the weight perpendicular to the incline. Since the incline is at an angle of 14.0 degrees, the normal force can be calculated as:
N = W * cos(14.0°) = 617.4 N * cos(14.0°) ≈ 591.3 N (perpendicular to the incline)
3. Force of friction:
The force of friction can be calculated using the formula: f = μ * N, where μ is the coefficient of kinetic friction and N is the normal force.
f = 0.27 * 591.3 N ≈ 159.6 N (opposite the direction of motion)
4. Tension force:
The tension force in the rope is acting parallel to the incline. It is given as 512 N.
Now, we can calculate the net force acting on the skier:
Net force = Tension force - Force of friction
Net force = 512 N - 159.6 N = 352.4 N
Next, we can calculate the acceleration using Newton's second law:
Net force = mass * acceleration
352.4 N = 63 kg * acceleration
Therefore, the acceleration of the skier is:
acceleration = 352.4 N / 63 kg ≈ 5.59 m/s²
Lastly, let's determine the direction of the skier's acceleration:
Since the net force is acting in the direction of the incline, the skier's acceleration is also in the direction of the incline (up the incline).
Therefore, the magnitude of the skier's acceleration is approximately 5.59 m/s², and the direction of the acceleration is up the incline.