3. An 80 kg skier is skiing down a slope (assume its frictionless) as shown in the diagram. The skier then hits a patch of sandy snow that applies a constant frictional force of 320 N directed up the slope.
a. What is the skiers acceleration before hitting the sandy snow?
b. Determine the skiers acceleration once they are on the sandy snow.
c. Assuming the skier started from rest 10 meters before the sandy patch, how far would the skier have to travel on the sandy snow to come to a rest?
To answer these questions, we need to use Newton's second law of motion, which states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass.
Given:
Mass of skier (m) = 80 kg
Frictional force before hitting sandy snow (F1) = 0 N (because it's a frictionless slope)
Frictional force on sandy snow (F2) = 320 N
Distance traveled before sandy snow (d1) = 10 m
a. To find the skier's acceleration before hitting the sandy snow, we can assume that there is no net force acting on the skier (since it's a frictionless slope). Therefore, the only force acting on the skier is their weight, which is calculated as the product of the mass and the acceleration due to gravity (9.8 m/s^2).
Weight of skier (W1) = mass × acceleration due to gravity = 80 kg × 9.8 m/s^2 = 784 N
Since there is no opposing force, the net force (F_net1) acting on the skier is equal to the weight.
F_net1 = W1 = 784 N
Using Newton's second law (F_net = m × a), we can rearrange the equation to solve for acceleration (a).
a = F_net1 / m = 784 N / 80 kg = 9.8 m/s^2
Therefore, the skier's acceleration before hitting the sandy snow is 9.8 m/s^2.
b. Once the skier is on the sandy snow, there is an additional force opposing their motion - the frictional force (F2) of 320 N. To find the skier's acceleration, we subtract the frictional force from the weight.
Net force on sandy snow (F_net2) = W1 - F2 = 784 N - 320 N = 464 N
Using the same equation as before, we can solve for acceleration (a).
a = F_net2 / m = 464 N / 80 kg = 5.8 m/s^2
Therefore, the skier's acceleration while on the sandy snow is 5.8 m/s^2.
c. To determine how far the skier would have to travel on the sandy snow to come to a rest, we need to use the equation of motion.
We know:
Initial velocity (u) = 0 m/s (since the skier started from rest)
Final velocity (v) = 0 m/s (since the skier comes to a rest)
Acceleration (a) = 5.8 m/s^2 (from part b)
The equation of motion is: v^2 = u^2 + 2as
Plugging in the values, we get:
0^2 = 0^2 + 2 × 5.8 m/s^2 × s
0 = 11.6 m/s^2 × s
Since the final velocity is zero, the distance traveled (s) will be the answer.
Therefore, the skier would have to travel a distance of 11.6 meters on the sandy snow to come to a rest.