Question

A 4.9-N hammer head is stopped from an initial downward velocity of 3.2 m/s in a distance of 0.45 cm by a nail in a pine board. In addition to its weight, there is a 15 N 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 F⃗ 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 only 0.12 cm. The downward forces on the hammer head are the same as on part (a). What then is the force F⃗ exerted by the hammer head on the nail while the hammer head is in contact with the nail and moving downward?

Answer 1

Answer:

a) -586 N

b) -2148 N

Explanation:

If the hammer head has a weight of 4.9 N

f = m * a

m = f / a

m = 4.9 / 9.8 = 0.5 kg

The hammer head has a mass of 0.5 kg.

It had a speed of 3.2 m/s when it hit the nail, and was completely stopped in 0.45 cm. During the stopping it had constant acceleration.

I set up a reference system with origin at the point where the hammer first hits the nail and the positive X axis pointing down.

Then I use the equation for movement onder cosntant acceleration.

X(t) = X0 + V0 * t + 1/2 * a * t^2

X0 = 0

X(t) = V0 + 1/2 * a * t^2

At X = 0.45 cm the velocity will be zero

The equation for velocity under constant acceleration is:

V(t) = V0 + a * t

0 = V0 + a * t

a * t = -V0

a = -V0 / t

Replacing

X(t) = V0 * t - 1/2 * V0 / t * t^2

0.0045 = 3.2 * t - 1.6 * t

0.0045 = 1.6 * t

t = 0.0045 / 1.6

t = 0.0028 s

a = -3.2 / 0.0028 = -1143 m/s^2

Then the total force on the hammer is of:

f = m * a

f = 0.5 * -1143 = -571 N

This force will be the resultant of the reaction force from the nail (negative) and the force applied y the person (+15 N)

ft = fn + fp

fn = ft - fp

fn = -571 - 15 = -586 N

If the distance traveled was 0.12 cm

0.0012 = 3.2 * t - 1.6 * t

0.0012 = 1.6 * t

t = 0.0012 / 1.6

t = 0.00075 s

a = -3.2 / 0.00075 = -4267 m/s^2

f = 0.5 * -4267 = -2133 N

ft = -2133 - 15 = -2148 N

Answer 2

In this exercise we have to use the knowledge of strength to calculate the strength does in certain situations, as follows:

a) -586 N

b) -2148 N

Organizing the information given in the statement, we have that:

• 4.9-N hammer
• velocity of 3.2 m/s
• distance of 0.45 cm
• 15 N downward force

Knowing the force formula is given by:

$f = m * a\\m = f / a\\m = 4.9 / 9.8 = 0.5 kg$

To find the acceleration we have to use the motion formula given as:

$X(t) = V0 * t - 1/2 * V0 / t * t^2\\0.0045 = 3.2 * t - 1.6 * t\\0.0045 = 1.6 * t\\t = 0.0045 / 1.6\\t = 0.0028 s\\a = -3.2 / 0.0028 = -1143 m/s^2$

A) Then the total force on the hammer is of:

$f = m * a\\f = 0.5 * -1143 = -571 N$

B) Now the other force will be:

$0.0012 = 3.2 * t - 1.6 * t\\0.0012 = 1.6 * t\\t = 0.0012 / 1.6\\t = 0.00075 s\\a = -3.2 / 0.00075 = -4267 m/s^2\\f = 0.5 * -4267 = -2133 N\\ft = -2133 - 15 = -2148 N$

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