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A nail in a pine board stops a 4.9-N hammer head from an initial downward velocity of 3.2 m/s in a distance of 0.45 cm. In addit
ion, the person using the hammer exerts a 15-N downward force on it. Assume that the acceleration of the hammer head is constant while it is in contact with the nail and moving downward.
(a) Draw a free-body diagram for the hammer head. Identify the reaction force for each action force in the diagram.
(b) Calculate the down-ward force F exerted by the hammer head on the nail while the hammer head is in contact with the nail and moving downward.
(c) Suppose that 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 in part
(d). What then 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?
(a) Draw a free-body diagram for the hammer head. Identify the reaction force for each action force in the diagram.
(b) Calculate the down-ward force F exerted by the hammer head on the nail while the hammer head is in contact with the nail and moving downward.
(c) Suppose that 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 in part
(d). What then 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?
1 answer:
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Answer:
Explanation:
According to the free-body diagram of the system, we have:
So, we can solve for T from (1):
Replacing (3) in (2):
The electric force () is given by the Coulomb's law. Recall that the charge q is the same in both spheres:
According to pythagoras theorem, the distance of separation (r) of the spheres are given by:
Finally, we replace (5) in (4) and solving for q:
Answer:
The maximum height the box will reach is 1.72 m
Explanation:
F = k·x
Where
F = Force of the spring
k = The spring constant = 300 N/m
x = Spring compression or stretch = 0.15 m
Therefore the force, F of the spring = 300 N/m×0.15 m = 45 N
Mass of box = 0.2 kg
Work, W, done by the spring = and the kinetic energy gained by the box is given by KE =
Since work done by the spring = kinetic energy gained by the box we have
=
therefore we have v =
=
=
= 5.81 m/s
Therefore the maximum height is given by
v² = 2·g·h or h = =
= 1.72 m