Answer:
It takes her 3.409 seconds to make a full stop.
Explanation:
The time it takes to make a full stop can be determined by the equation of velocity for a Uniformly Accelerated Rectilinear Motion:
(1)
Where is the final velocity,
is the initial velocity, a is the acceleration and t is the time.
Equation (1) can be rewritten in terms of t:
(2)
For this particular case the final velocity will be equal to zero ():
So it takes her 3.409 seconds to make a full stop.
(a) 69.3 J
The work done by the applied force is given by:
where:
F = 25.0 N is the magnitude of the applied force
d = 3.20 m is the displacement of the sled
is the angle between the direction of the force and the displacement of the sled
Substituting numbers into the formula, we find
(b) 0
The problem says that the surface is frictionless: this means that no friction is acting on the sled, therefore the energy dissipated by friction must be zero.
(c) 69.3 J
According to the work-energy theorem, the work done by the applied force is equal to the change in kinetic energy of the sled:
where
is the change in kinetic energy
W is the work done
Since we already calculated W in part (a):
W = 69.3 J
We therefore know that the change in kinetic energy of the sled is equal to this value:
(d) 4.9 m/s
The change in kinetic energy of the sled can be rewritten as:
(1)
where
Kf is the final kinetic energy
Ki is the initial kinetic energy
m = 5.50 kg is the mass of the sled
u = 0 is the initial speed of the sled
v = ? is the final speed of the sled
We can calculate the variation of kinetic energy of the sled, , after it has travelled for d=3 m. Using the work-energy theorem again, we find
And substituting into (1) and re-arrangin the equation, we find
A. 6.67 cm
The focal length of the lens can be found by using the lens equation:
where we have
f = focal length
s = 12 cm is the distance of the object from the lens
s' = 15 cm is the distance of the image from the lens
Solving the equation for f, we find
B. Converging
According to sign convention for lenses, we have:
- Converging (convex) lenses have focal length with positive sign
- Diverging (concave) lenses have focal length with negative sign
In this case, the focal length of the lens is positive, so the lens is a converging lens.
C. -1.25
The magnification of the lens is given by
where
s' = 15 cm is the distance of the image from the lens
s = 12 cm is the distance of the object from the lens
Substituting into the equation, we find
D. Real and inverted
The magnification equation can be also rewritten as
where
y' is the size of the image
y is the size of the object
Re-arranging it, we have
Since in this case M is negative, it means that y' has opposite sign compared to y: this means that the image is inverted.
Also, the sign of s' tells us if the image is real of virtual. In fact:
- s' is positive: image is real
- s' is negative: image is virtual
In this case, s' is positive, so the image is real.
E. Virtual
In this case, the magnification is 5/9, so we have
which can be rewritten as
which means that s' has opposite sign than s: therefore, the image is virtual.
F. 12.0 cm
From the magnification equation, we can write
and then we can substitute it into the lens equation:
and we can solve for s:
G. -6.67 cm
Now the image distance can be directly found by using again the magnification equation:
And the sign of s' (negative) also tells us that the image is virtual.
H. -24.0 cm
In this case, the image is twice as tall as the object, so the magnification is
M = 2
and the distance of the image from the lens is
s' = -24 cm
The problem is asking us for the image distance: however, this is already given by the problem,
s' = -24 cm
so, this is the answer. And the fact that its sign is negative tells us that the image is virtual.