Now let's apply the work-energy theorem to a more complex, multistep problem. In a pile driver, a steel hammerhead with mass 200 kg is lifted 3.00 m above the top of a vertical I-beam that is to be driven into the ground (Figure 1). The hammer is then dropped, driving the I-beam 7.40 cm farther into the ground. The vertical rails that guide the hammerhead exert a constant 60.0 N friction force on it. Use the work-energy theorem to find

(a) the speed of the hammerhead just as it hits the I-beam and

(b) the average force the hammerhead exerts on the I-beam.

Question

Hurley

Physics

9 September, 17:13

Now let's apply the work-energy theorem to a more complex, multistep problem. In a pile driver, a steel hammerhead with mass 200 kg is lifted 3.00 m above the top of a vertical I-beam that is to be driven into the ground (Figure 1). The hammer is then dropped, driving the I-beam 7.40 cm farther into the ground. The vertical rails that guide the hammerhead exert a constant 60.0 N friction force on it. Use the work-energy theorem to find

(a) the speed of the hammerhead just as it hits the I-beam and

(b) the average force the hammerhead exerts on the I-beam.

+3

Answers (1)

Know the Answer?

Giuliana Brennan · Answer 1

a) v = 7.67

b) n = 81562 N

Explanation:

Given:-

- The mass of hammer-head, m = 200 kg

- The height at from which hammer head drops, s12 = 3.00 m

- The amount of distance the I-beam is hammered, s23 = 7.40 cm

- The resistive force by contact of hammer-head and I-beam, F = 60.0 N

Find:-

(a) the speed of the hammerhead just as it hits the I-beam and

(b) the average force the hammerhead exerts on the I-beam.

Solution:-

- We will consider the hammer head as our system and apply the conservation of energy principle because during the journey of hammer-head up till just before it hits the I-beam there are no external forces acting on the system:

ΔK. E = ΔP. E

K_2 - K_1 = P_1 - P_2

Where, K_2: Kinetic energy of hammer head as it hits the I-beam

K_1: Initial kinetic energy of hammer head ( = 0) ... rest

P_2: Gravitational potential energy of hammer head as it hits the I-beam. (Datum = 0)

P_1: Initial gravitational potential energy of hammer head

- The expression simplifies to:

K_2 = P_1

Where, 0.5*m*v2^2 = m*g*s12

v2 = √ (2*g*s12) = √ (2*9.81*3)

v2 = 7.67 m/s

- For the complete journey we see that there are fictitious force due to contact between hammer-head and I-beam the system is no longer conserved. All the kinetic energy is used to drive the I-beam down by distance s23. We will apply work energy principle on the system:

Wnet = (P_3 - P_1) + W_friction

Wnet = m*g*s13 + F*s23

n*s23 = m*g*s13 + F*s23

Where, n: average force the hammerhead exerts on the I-beam.

s13 = s12 + s23

Hence,

n = m*g * (s12/s23 + 1) + F

n = 200*9.81 * (3/0.074 + 1) + 60

n = 81562 N