A hiker throws a rock with a mass of 1.4kg from a cliff that is 37 m above the ground . The hiker gives the rock an initial downward velocity of 8.5 m/s.
a) What is the energy of the rock at the moment the hiker releases it?
b) What is the velocity of the rock at the moment it hits the ground?
c) What velocity would the hiker have had to give the rock in order for it to have 3.6 kJ of energy when it hits the ground?
a.KE = 0.5m*V^2 = 0.5*1.4*8.5^2=50.6 J.
b. V^2 = Vo^2 + 2g*d
V^2 = 8.5^2 + 19.6*37 = 725.2
V = 26.9 m/s.
c. 0.5m*V^2 = 3600 J.
.5*1.4*V^2 = 3600
0.7V^2 = 3600
V^2 = 5143
V = 71.7 m/s. When it hits the gnd.
V^2 = Vo^2 + 2g*h
Vo^2 = V^2-2g*h
Vo^2 = (71.7)^2 - 19.6*37 = 4416
Vo = 66.5 m/s. Gives the rock 3.6 kJ of
energy.
To find the answers to these questions, we need to apply the principles of energy and motion.
a) What is the energy of the rock at the moment the hiker releases it?
The energy of the rock at the moment it is released can be calculated using the concept of gravitational potential energy. Gravitational potential energy can be calculated using the formula:
Potential Energy = mass × gravitational acceleration × height
In this case, the mass of the rock is 1.4 kg, the height is 37 m, and the gravitational acceleration is approximately 9.8 m/s². Plugging these values into the formula, we get:
Potential Energy = 1.4 kg × 9.8 m/s² × 37 m
Simplifying this expression, we find that the potential energy of the rock at the moment of release is 508.04 J (Joules).
b) What is the velocity of the rock at the moment it hits the ground?
To calculate the velocity of the rock at the moment it hits the ground, we can use the principle of conservation of energy. At the moment of impact, all the potential energy of the rock is converted into kinetic energy. The kinetic energy formula is given by:
Kinetic Energy = (1/2) × mass × velocity²
Since the potential energy of the rock is converted to kinetic energy at the moment it hits the ground, we set the potential energy equal to the kinetic energy:
Potential Energy = Kinetic Energy
Substituting the values we have, we get:
508.04 J = (1/2) × 1.4 kg × velocity²
Now we can solve for the velocity. Rearranging the equation, we find that:
velocity² = (2 × 508.04 J) / 1.4 kg
velocity² = 727.2 m²/s²
Taking the square root of both sides, we get:
velocity = √727.2 m/s
Thus, the velocity of the rock at the moment it hits the ground is approximately 26.98 m/s.
c) What velocity would the hiker have had to give the rock in order for it to have 3.6 kJ of energy when it hits the ground?
To calculate the velocity required for the rock to have a specific energy when it hits the ground, we can use the same principle of conservation of energy. In this case, the energy of the rock is given as 3.6 kJ (kilojoules), which is equal to 3,600 J.
Setting this energy equal to the kinetic energy of the rock, we have:
3,600 J = (1/2) × 1.4 kg × velocity²
Simplifying this equation, we find:
velocity² = (2 × 3,600 J) / 1.4 kg
velocity² = 5,142.86 m²/s²
Taking the square root of both sides, we get:
velocity = √5,142.86 m/s
Therefore, the hiker would have had to give the rock a velocity of approximately 71.74 m/s for it to have 3.6 kJ of energy when it hits the ground.