A sight seen on many bunny hills across Ontario is young skiers pushing on ski poles and gliding down a slope until they come to rest. Observing from a distance, you note a young person (approximately 25 kg) pushing off with the ski poles to give herself an initial velocity of 3.5 m/s. If the inclination of the hill is 5.0 degrees and the coefficient of kinetic friction for the skis on dry snow is 0.20, calculate
a) the time taken for the skier to come to a stop
b) the distance travelled down the hill
component of weight down slope = m g sin 5
normal force = m g cos 5 so friction up slope = .2 m g cos 5
F = net force down slope = m g (sin 5 - .2 cos 5)
= m a
so a down slope = g(sin 5 - .2 cos 5) = g (-0.112)
if g = 9.81
then a = - 1.1 m/s^2
now the problem
v = Vi - 1.1 t = 3.5 -1.1 t
when v =0
t = 3.5/1.1 = 3.18 seconds
distance = average speed * t = (3.5/2) * 3.18 = 5.57 meters
thank you Damon
To calculate the time taken for the skier to come to a stop, we can use the concept of mechanical work and energy.
a) First, let's calculate the gravitational force acting on the skier. The gravitational force can be calculated using the formula:
F_grav = m * g
where m is the mass of the skier (25 kg) and g is the acceleration due to gravity (9.8 m/s^2). Substituting the values, we have:
F_grav = 25 kg * 9.8 m/s^2 = 245 N
Next, let's calculate the force opposing the motion, which is the force of kinetic friction. The force of kinetic friction can be calculated using the formula:
F_friction = u * F_normal
where u is the coefficient of kinetic friction (0.20) and F_normal is the normal force acting on the skier. The normal force can be calculated using the formula:
F_normal = m * g * cos(theta)
where theta is the angle of inclination of the hill (5.0 degrees). Converting the angle to radians:
theta = 5.0 degrees * (pi/180) = 0.08726 radians
Substituting the values, we have:
F_normal = 25 kg * 9.8 m/s^2 * cos(0.08726) = 242.9 N
Substituting the values of u and F_normal, we can find the force of kinetic friction:
F_friction = 0.20 * 242.9 N = 48.58 N
Now, let's calculate the net force acting on the skier. The net force can be calculated using the formula:
F_net = F_push - F_friction
where F_push is the force exerted by the skier initially (impulse) to give herself an initial velocity of 3.5 m/s. Since the skier is at rest initially, the net force is equal to the force of kinetic friction. Therefore:
F_friction = F_push
Simplifying the equation, we have:
F_push = 48.58 N
Next, let's calculate the acceleration of the skier using Newton's second law of motion:
F_push = m * a
Substituting the known values, we have:
48.58 N = 25 kg * a
Solving for a, we get:
a = 48.58 N / 25 kg = 1.9432 m/s^2
Now, let's calculate the time taken for the skier to come to a stop using the equation of motion:
v = u + a * t
where v is the final velocity of the skier (0 m/s), u is the initial velocity of the skier (3.5 m/s), a is the acceleration of the skier (-1.9432 m/s^2), and t is the time taken. Substituting the values, we have:
0 m/s = 3.5 m/s + (-1.9432 m/s^2) * t
Solving for t, we get:
t = (0 m/s - 3.5 m/s) / (-1.9432 m/s^2) ≈ 1.799 seconds
Therefore, the time taken for the skier to come to a stop is approximately 1.799 seconds.
b) To calculate the distance travelled down the hill, we can use the equation of motion:
s = ut + (1/2) * a * t^2
where s is the distance traveled down the hill, u is the initial velocity of the skier (3.5 m/s), a is the acceleration of the skier (-1.9432 m/s^2), and t is the time taken (1.799 seconds). Substituting the values, we have:
s = (3.5 m/s) * (1.799 s) + (1/2) * (-1.9432 m/s^2) * (1.799 s)^2
Simplifying the equation, we get:
s ≈ 2.668 meters
Therefore, the distance traveled down the hill is approximately 2.668 meters.