Answer:
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
