IDENTIFY: We’ll use the work–energy theorem to relate the hammerhead’s speed at different locations and the forces acting on it.
There are three locations of interest: point 1, where the hammerhead starts from rest; point 2, where it first contacts the I-beam; and point 3, where the hammerhead and I-beam come to a halt (Fig. 6.12a). The two target variables are the hammerhead’s speed at point 2 and the average force the hammerhead exerts between points 2 and 3. Hence we’ll apply the work–energy theorem twice: once for the motion from 1 to 2, and once for the motion from 2 to 3.
SET UP: Figure 6.12b shows the vertical forces on the hammerhead as it falls from point 1 to point 2. (We can ignore any horizontal forces that may be present because they do no work as the hammerhead moves vertically.) For this part of the motion, our target variable is the hammerhead’s final speed v_2.
Figure 6.12c shows the vertical forces on the hammerhead during the motion from point 2 to point 3. In addition to the forces shown in Fig. 6.12b, the I-beam exerts an upward normal force of magnitude n on the hammerhead. This force actually varies as the hammerhead comes to a halt, but for simplicity we’ll treat n as a constant. Hence n represents the average value of this upward force during the motion. Our target variable for this part of the motion is the force that the hammerhead exerts on the I-beam; it is the reaction force to the normal force exerted by the I-beam, so by Newton’s third law its magnitude is also n.
EXECUTE:
(a) From point 1 to point 2, the vertical forces are the downward weight w = mg = (200 kg)(9.8 m/s^2) = 1960 N and the upward friction force f = 60 N. Thus the net downward force is w – f = 1900 N. The displacement of the hammerhead from point 1 to point 2 is downward and equal to s_{12} = 3.00 m. The total work done on the hammerhead between point 1 and point 2 is then
W_{\mathrm{tot}}=(w-f) s_{12}=(1900 \mathrm{~N})(3.00 \mathrm{~m})=5700 \mathrm{~J}
At point 1 the hammerhead is at rest, so its initial kinetic energy K_1 is zero. Hence the kinetic energy K_2 at point 2 equals the total work done on the hammerhead between points 1 and 2:
W_{\mathrm{tot}}=K_{2}-K_{1}=K_{2}-0=\frac{1}{2} m v_{2}^{2}-0
v_{2}=\sqrt{\frac{2 W_{\mathrm{tot}}}{m}}=\sqrt{\frac{2(5700 \mathrm{~J})}{200 \mathrm{~kg}}}=7.55 \mathrm{~m} / \mathrm{s}
This is the hammerhead’s speed at point 2, just as it hits the I-beam.
(b) As the hammerhead moves downward from point 2 to point 3, its displacement is s_{23} = 7.4 cm = 0.074 m and the net downward force acting on it is w – f – n (Fig. 6.12c). The total work done on the hammerhead during this displacement is
W_{\mathrm{tot}}=(w-f-n) s_{23}
The initial kinetic energy for this part of the motion is K_2, which from part (a) equals 5700 J. The final kinetic energy is K_3 = 0 (the hammerhead ends at rest). From the work–energy theorem,
W_{\mathrm{tot}}=(w-f-n) s_{23}=K_{3}-K_{2}
n=w-f-\frac{K_{3}-K_{2}}{s_{23}}=1960 \mathrm{~N}-60 \mathrm{~N}-\frac{0 \mathrm{~J}-5700 \mathrm{~J}}{0.074 \mathrm{~m}}=79,000 \mathrm{~N}
The downward force that the hammerhead exerts on the I-beam has this same magnitude, 79,000 N (about 9 tons)—more than 40 times the weight of the hammerhead.
EVALUATE: The net change in the hammerhead’s kinetic energy from point 1 to point 3 is zero; a relatively small net force does positive work over a large distance, and then a much larger net force does negative work over a much smaller distance. The same thing happens if you speed up your car gradually and then drive it into a brick wall. The very large force needed to reduce the kinetic energy to zero over a short distance is what does the damage to your car—and possibly to you.
