A skier of mass 109 kg travels down a frictionless ski trail.
a)If the top of the trail is a height 218 m above the bottom, what is the work done by gravity on the skier? _____J
b)Find the velocity of the skier when he reaches the bottom of the ski trail. Assume he starts from rest. ____m/s
a) The work done by gravity on an object can be calculated using the formula: W = mgh, where W is the work done, m is the mass of the object, g is the acceleration due to gravity, and h is the height.
In this case, the mass of the skier is given as 109 kg and the height is 218 m. The acceleration due to gravity is approximately 9.8 m/s^2.
Using the formula, we can calculate the work done by gravity:
W = (109 kg) * (9.8 m/s^2) * (218 m)
W ≈ 241,628 J
Therefore, the work done by gravity on the skier is approximately 241,628 J.
b) To find the velocity of the skier when he reaches the bottom of the ski trail, we can use the principle of conservation of energy. The initial potential energy at the top of the trail will be converted into kinetic energy at the bottom.
The potential energy at the top of the trail can be calculated using the formula: PE = mgh, where PE is the potential energy, m is the mass, g is the acceleration due to gravity, and h is the height.
Using the given values, the potential energy at the top of the trail is:
PE = (109 kg) * (9.8 m/s^2) * (218 m)
PE ≈ 233,032 J
Since there is no friction, this potential energy will be completely converted into kinetic energy at the bottom of the trail.
Kinetic energy (KE) can be calculated using the formula: KE = 0.5 * mv^2, where KE is the kinetic energy, m is the mass, and v is the velocity.
Setting the potential energy equal to the kinetic energy:
233,032 J = 0.5 * (109 kg) * v^2
Simplifying the equation:
v^2 = (2 * 233,032 J) / (109 kg)
v^2 ≈ 4271.87 m^2/s^2
Taking the square root of both sides:
v ≈ √(4271.87 m^2/s^2)
v ≈ 65.37 m/s
Therefore, the velocity of the skier when he reaches the bottom of the ski trail is approximately 65.37 m/s.
a) To find the work done by gravity on the skier, we can use the equation:
Work = Force x Distance
The only force acting on the skier is the force of gravity, which can be calculated using the equation:
Force of gravity = mass x acceleration due to gravity
In this case, the mass of the skier is given as 109 kg, and the acceleration due to gravity is approximately 9.8 m/s².
So, the force of gravity acting on the skier is:
Force of gravity = 109 kg x 9.8 m/s² = 1068.2 N (rounded to one decimal place)
Now, we need to calculate the distance traveled by the skier. The distance is given as the height of the trail, which is 218 m.
Finally, we can substitute these values into the work formula to find the work done by gravity:
Work = 1068.2 N x 218 m = 232,533.6 J (rounded to one decimal place)
Therefore, the work done by gravity on the skier is approximately 232,533.6 J.
b) To find the velocity of the skier when he reaches the bottom of the ski trail, we can use the principle of conservation of mechanical energy. Since there is no friction and the ski trail is assumed to be frictionless, the total mechanical energy of the skier remains constant throughout the descent.
The total mechanical energy can be calculated using the equation:
Total mechanical energy = Potential energy + Kinetic energy
At the top of the trail, the skier only has potential energy, given by:
Potential energy = mass x gravitational acceleration x height
So, at the top of the trail:
Potential energy = 109 kg x 9.8 m/s² x 218 m = 234,748.4 J (rounded to one decimal place)
When the skier reaches the bottom of the trail, all the potential energy is now converted into kinetic energy:
Kinetic energy = (1/2) x mass x velocity^2
Since the skier starts from rest (zero velocity), the kinetic energy at the bottom is also zero.
Equating the initial potential energy to the final kinetic energy:
Potential energy = Kinetic energy
234,748.4 J = (1/2) x 109 kg x velocity^2
Simplifying:
velocity^2 = (234,748.4 J x 2) / (109 kg)
velocity^2 = 4,317.9321 m²/s²
Taking the square root of both sides:
velocity = √(4,317.9321 m²/s²) = 65.7 m/s (rounded to one decimal place)
Therefore, the velocity of the skier when he reaches the bottom of the ski trail is approximately 65.7 m/s.