A drop hammer of mass 120 kg falls 2.5 m onto a pile of mass 250 kg and drives it 70 mm into the ground. Calculate:
a.The velocity with which the hammer strikes the pile (use the conservation of energy principle)
b.The velocity immediately after the impact(use the conservation of momentum principle,
assuming the hammer does not rebound on impact)
c. The average ground resistance (use work-energy method)
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Sure! Let's solve each part step by step:
a. To find the velocity with which the hammer strikes the pile, we can use the principle of conservation of energy. The initial potential energy of the hammer is equal to the kinetic energy just before it strikes the pile.
The potential energy (PE) of the hammer is given by the formula:
PE = mgh
Where:
m = mass of the hammer = 120 kg
g = acceleration due to gravity = 9.8 m/s^2
h = height through which the hammer falls = 2.5 m
So, the potential energy of the hammer is:
PE = 120 kg * 9.8 m/s^2 * 2.5 m = 2940 J
This potential energy is converted into kinetic energy just before the hammer strikes the pile.
The kinetic energy (KE) of an object is given by the formula:
KE = 1/2 * mv^2
Where:
m = mass of the hammer = 120 kg
v = velocity of the hammer just before it strikes the pile (to be determined)
Since the potential energy is converted completely into kinetic energy, we can equate the two:
PE = KE
2940 J = 1/2 * 120 kg * v^2
Solving for v, we get:
v^2 = (2 * 2940 J) / 120 kg
v^2 = 49
v ≈ √49 = 7 m/s
Therefore, the velocity with which the hammer strikes the pile is approximately 7 m/s.
b. To find the velocity immediately after the impact, we can use the principle of conservation of momentum. The total momentum before the impact is equal to the total momentum after the impact.
The momentum (p) of an object is given by the formula:
p = mv
Initial momentum before the impact:
p1 = momentum of the hammer = mass of the hammer * velocity of the hammer just before it strikes the pile
p1 = 120 kg * 7 m/s = 840 kg·m/s
Final momentum after the impact:
p2 = momentum of the combined system (hammer + pile) just after the impact
p2 = (mass of the hammer + mass of the pile) * velocity of the combined system just after the impact (to be determined)
Since the hammer does not rebound on impact, we can assume that the final velocity of the combined system is the same as the velocity of the pile just after the impact.
Applying the principle of conservation of momentum, we have:
p1 = p2
840 kg·m/s = (120 kg + 250 kg) * v2
Solving for v2, we get:
v2 = 840 kg·m/s / (120 kg + 250 kg)
v2 ≈ 840 kg·m/s / 370 kg
v2 ≈ 2.27 m/s
Therefore, the velocity immediately after the impact is approximately 2.27 m/s.
c. To find the average ground resistance using the work-energy method, we can calculate the work done by the ground to stop the pile.
The work done (W) is given by the formula:
W = force * distance
The force (F) required to stop the pile is equal to the average ground resistance.
The distance (d) through which the pile is driven into the ground is given as 70 mm which is equal to 0.07 m.
The work done by the ground is equal to the change in kinetic energy of the pile.
The initial kinetic energy (KE1) is zero since the pile is stationary.
The final kinetic energy (KE2) is given by the formula:
KE2 = 1/2 * mv^2
Where:
m = mass of the pile = 250 kg
v = velocity of the pile immediately after the impact = 2.27 m/s (from part b)
The work done by the ground can be calculated as:
W = KE2 - KE1
W = 1/2 * 250 kg * (2.27 m/s)^2 - 0
W = 1/2 * 250 kg * (2.27 m/s)^2
W ≈ 645 J
Therefore, the average ground resistance is approximately 645 J.