A man throws a rock of mass m = 0.310 kg straight up into the air. In this process, his arm does a total amount of work Wnet = 105 J on the rock. Calculate the maximum distance, h, above the man's throwing hand that the rock will travel. Neglect air resistance.
Wnet = m g H
Solve for the maximum height reached, H.
Is Wnet 105 or 10^5 joules? 105 is more likely
Yes, Wnet is 105 J.
Thank you.
To calculate the maximum distance above the man's throwing hand that the rock will travel, we can use the principle of conservation of energy. Since there is no air resistance, the only energies involved in this scenario are kinetic energy and gravitational potential energy.
Let's calculate the kinetic energy of the rock when it leaves the man's hand. The kinetic energy (KE) is given by:
KE = 0.5 * mass * velocity^2
Since the rock is thrown straight up, its velocity at the maximum height will be zero. Therefore, the initial kinetic energy is equal to the net work done on the rock:
KE_initial = W_net
Substituting the given value W_net = 105 J, we have:
KE_initial = 105 J
Next, let's calculate the gravitational potential energy (PE) at the maximum height. The potential energy is given by:
PE = mass * g * height
Where g is the acceleration due to gravity, which is approximately 9.8 m/s^2 on Earth.
At the maximum height, the rock will have the maximum potential energy but zero kinetic energy. Therefore, we have:
PE_max = KE_initial
Solving for the height, we get:
mass * g * height = KE_initial
Substituting the given values mass = 0.310 kg and g = 9.8 m/s^2, we have:
0.310 kg * 9.8 m/s^2 * height = 105 J
Now, solve for the height:
height = 105 J / (0.310 kg * 9.8 m/s^2)
Calculating this value gives us:
height ≈ 35.664 m
Therefore, the maximum distance above the man's throwing hand that the rock will travel is approximately 35.664 meters.