Suppose you hit a steel nail with a 0.500-kg hammer, initially moving at 15.0 m/s and brought to rest in 2.80 mm. (a) What average force is exerted on the nail? (b) How much is the nail compressed if it is 2.50 mm in diameter and 6.00-cm long? (c) What pressure is created on the 1.00-mmdiameter tip of the nail?

Step 1 of 9</p>

a.)

The average force exerted on the nail can be computed by equating the kinetic of energy of the hammer to work done in bringing the nail to rest at 2.8 mm

Step 2 of 9

The kinetic energy of the hammer is given by the expression,

where, m = mass of the hammer in kg

v = velocity of the hammer in m/s

Substituting 0.500 kg for m and 15.0 m/s for v

=

= 56.25 J

Step 3 of 9</p>

Work done in bringing the nail to rest can be calculated as

where, F = Force exerted on the nail in N

d= distance travelled by the nail in m

Substituting 2.810-2 m for d,

Step 4 of 9</p>

The average force exerted on the nail can be now determined by equating the kinetic energy of the hammer to work done in bringing the nail to rest.

Solving for F,

= 2008.93 N

Therefore, the average force exerted on the nail is 2008.93 N

b.)

Step 5 of 9

The compression l of the nail can be found by using the relation between the Stress, Strain and Young’s modulus for the steel nail.

Young’s modulus =

or

where, F = Force applied in N

A = Area on which force applied in m2

l = original length of the nail in m

l = change in the length of the nail (compression) in m

= Young’s modulus of the steel = 180109 N/m2

Solving for l,