# Hammer problem: Newton's Laws

• Jul 24th 2007, 04:42 PM
Hammer problem: Newton's Laws
A 4.9N hammer head is stopped from an initial downward velocity of 3.0m/s in a distance of 0.45cm by a nail in a pine board. In addition to its weight, there is a 15N downward force on the hammer head applied by the person using the hammer. Assume that the acceleration of the hammer head is constant while it is in contact with the nail and moving downward.

a.)Calculate the downward force F exerted by the hammer head on the nail while the hammer head is in contact with the nail and moving downward.

b.)Suppose the nail is in hardwood and the distance the hammer head travels in coming to rest is 0.15cm. The downward force on the hammer head is the same as in part a.) What is the force F exerted by the hammer head on the nail while the hammer head is in contact with the nail and moving downward?

need help with physics pros here the answer is 1.00kN and 3.00kN
• Jul 25th 2007, 05:28 AM
topsquark
Quote:

A 4.9N hammer head is stopped from an initial downward velocity of 3.0m/s in a distance of 0.45cm by a nail in a pine board. In addition to its weight, there is a 15N downward force on the hammer head applied by the person using the hammer. Assume that the acceleration of the hammer head is constant while it is in contact with the nail and moving downward.

a.)Calculate the downward force F exerted by the hammer head on the nail while the hammer head is in contact with the nail and moving downward.

b.)Suppose the nail is in hardwood and the distance the hammer head travels in coming to rest is 0.15cm. The downward force on the hammer head is the same as in part a.) What is the force F exerted by the hammer head on the nail while the hammer head is in contact with the nail and moving downward?

need help with physics pros here the answer is 1.00kN and 3.00kN

The force applied by the person is extra information here. Ignore it.

Technically you aren't looking for Newton's 2nd Law here, exactly, but another form of it known as the "Impulse-Momentum" theorem. The specific form you want is:
$\bar{F} \Delta t = m \Delta v$
where $\bar{F}$ is the average force on the hammer and $\Delta t$ is the time during which the hammer is decelerating.

So $\bar{F} = m \frac{\Delta v}{\Delta t} = m \bar{a}$
where $\bar{a}$ is the average acceleration.

I disagree with both the given answers here.

Part a) We know that the nail moves a distance of 0.45 cm when in contact with the hammer, so this is the distance the hammer moved from the time of contact to when the hammer stopped (assuming the hammer doesn't rebound, which it should.) We know the initial speed of the hammer, so assuming a constant deceleration, we can find the deceleration:
$v^2 = v_0^2 + 2a(x - x_0)$
(For a coordinate system here I am taking as the origin the point where the hammer first strikes the nail and a +x direction in the direction that the hammer travels while "pounding" the nail.)
$0^2 = 3^2 + 2a(0.0045 - 0)$ <-- Dropping the units for convenience

$a = -\frac{3^2}{2 \cdot 0.0045} = -1000 ~m/s^2$

Taking the magnitude of this I get that
$\bar{F} = m \bar{a} = \left ( \frac{4.9~N}{9.8~m/s^2} \right ) (1000~m/s^2) = 500~N$

Part b) is done in exactly the same way and I get that
$\bar{F} = 1500~N$

I note that the accelerations are 1000 m/s^2 and 3000 m/s^2, respectively, which are numerically the answers you listed. Perhaps someone copied the wrong numbers when giving you the answers?

-Dan
• Jul 25th 2007, 06:23 AM