When a 4.60-kg object is hung vertically on a certain light spring that obeys Hooke's law, the spring stretches 2.30 cm. (a) If
the 4.60-kg object is removed, how far will the spring stretch if a 1.50-kg block is hung on it? cm (b) How much work must an external agent do to stretch the same spring 4.00 cm from its unstretched position?
1 answer:
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Answer:
Fe= 28.2 N : Magnitude of the equilibrant (Fe)
β = 18.34° , clockwise from the positive x axis
Explanation:
Concept of the equilibrant
It is called equilibrant to a force with the same magnitude and direction as the resulting one (in case it is non-zero) but in the opposite direction. Adding vectorially to all the forces (that is to say the resulting one) with the equilibrant you get zero
To solve this problem we decompose the forces given into x-y components to find the resulting force:
Look at the attached graphic
F₁= 33.4 N , θ₁=23.8° clockwise from the positive y axis (y+)
F₁x= 33.4 *sin23.8° = 13.48 N
F₁y= 33.4 *cos23.8° =30.6 N
F₂=46.1 N , θ₂=28.8 counterclockwise from the negative x axis (x-)
F₂x= -46.1 *cos28.8° = -40.4 N
F₂y= -46.1 *sin28.8° = -22.2 N
Components of the resultant in x-y R(x,y)
Rx= 13.48 N -40.4 N = - 26.92 N
Ry= 30.6 N -22.2 N = + 8.4 N
Components of the equilibrant in x-y Fe(x,y)
Fex= +26.92 N
Fey= - 8.4 N
Magnitude of the equilibrant (Fe)
Fe= 28.2 N
Angle the equilibrant makes with the x axis ( β)
β = -18.34°
β = 18.34° , clockwise from the positive x axis
A) The motion of the ball consists of two indipendent motions on the horizontal (x) and vertical (y) axis. The laws of motion in the two directions are:
where
- the horizontal motion is a uniform motion, with constant speed
, where 
and 
- the vertical motion is an uniformly accelerated motion, with constant acceleration
, initial position h (the height of the building) and initial vertical velocity 
(with a negative sign, since it points downwards)
The ball strikes the ground after a time t=3.00 s, so we can find the distance covered horizontally by the ball by substituting t=3.00 s into the equation of x(t):
b) To find the height from which the ball was thrown, h, we must substitute t=3.00 s into the equation of y(t), and requiring that y(3.00 s)=0 (in fact, after 3 seconds the ball reaches the ground, so its vertical position y(t) is zero). Therefore, we have:
which becomes
c) We want the ball to reach a point 10.0 meters below the level of launching, so we want to find the time t such that
If we substitute this into the equation of y(t), we have
whose solution is
(the other solution is negative, so it has no physical meaning). Therefore, the ball reaches a point 10 meters below the level of launching after 1.18 s.
where
- the horizontal motion is a uniform motion, with constant speed
- the vertical motion is an uniformly accelerated motion, with constant acceleration
The ball strikes the ground after a time t=3.00 s, so we can find the distance covered horizontally by the ball by substituting t=3.00 s into the equation of x(t):
b) To find the height from which the ball was thrown, h, we must substitute t=3.00 s into the equation of y(t), and requiring that y(3.00 s)=0 (in fact, after 3 seconds the ball reaches the ground, so its vertical position y(t) is zero). Therefore, we have:
which becomes
c) We want the ball to reach a point 10.0 meters below the level of launching, so we want to find the time t such that
If we substitute this into the equation of y(t), we have
whose solution is