A force of 157 N is applied to a box 25o above the horizontal. If it is applied over a distance of 14.5 m how much work is done?
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Before the engines fail, the rocket's altitude at time <em>t</em> is given by
and its velocity is
The rocket then reaches an altitude of 1150 m at time <em>t</em> such that
Solve for <em>t</em> to find this time to be
At this time, the rocket attains a velocity of
When it's in freefall, the rocket's altitude is given by
where is the acceleration due to gravity, and its velocity is
(a) After the first 11.2 s of flight, the rocket is in the air for as long as it takes for to reach 0:
So the rocket is in motion for a total of 11.2 s + 32.6 s = 43.4 s.
(b) Recall that
where and
denote final and initial velocities, respecitively,
denotes acceleration, and
the difference in altitudes over some time interval. At its maximum height, the rocket has zero velocity. After the engines fail, the rocket will keep moving upward for a little while before it starts to fall to the ground, which means
will contain the information we need to find the maximum height.
Solve for and we find that the rocket reaches a maximum altitude of about 1930 m.
(c) In part (a), we found the time it takes for the rocket to hit the ground (relative to ) to be about 32.6 s. Plug this into
to find the velocity before it crashes:
That is, the rocket has a velocity of 196 m/s in the downward direction as it hits the ground.
To solve this problem we will apply the concepts related to equilibrium, for this specific case, through the sum of torques.
If the distance in which the 600lb are applied is 6in, we will have to add the unknown Force sum, at a distance of 27in - 6in will be equivalent to that required to move the object. So,
So, Force that must be applied at the long end in order to lift a 600lb object to the short end is 171.42lb