Suppose the spring begins in a compressed state, so that the block speeds up from rest to 2.6 m/s as it passes through the equilibrium point, and so that when it first comes to a stop, the spring is stretched 0.20 m.
There are two forces performing work on the block: the restoring force of the spring and kinetic friction.
By the work-energy theorem, the total work done on the block between the equilbrium point and the 0.20 m mark is equal to the block's change in kinetic energy:

or

where <em>K</em> is the block's kinetic energy at the equilibrium point,

Both the work done by the spring and by friction are negative because these forces point in the direction opposite the block's displacement. The work done by the spring on the block as it reaches the 0.20 m mark is

Compute the work performed by friction:

By Newton's second law, the net vertical force on the block is
∑ <em>F</em> = <em>n</em> - <em>mg</em> = 0 ==> <em>n</em> = <em>mg</em>
where <em>n</em> is the magnitude of the normal force from the surface pushing up on the block. Then if <em>f</em> is the magnitude of kinetic friction, we have <em>f</em> = <em>µmg</em>, where <em>µ</em> is the coefficient of kinetic friction.
So we have



Answer:
i)4080000V
ii)4080000V
iii)5766400v
Explanation:
Hello!
To solve this problem we will use the following steps
1. We will use the equation that defines the distance between two points in order to find the distance from the origin to each of the points that the problem asks for.

x= horizonalt component
y=vertical component
d=distance
i)
x=0
y=6

ii)
x=6
y=0

iii).
x=6
y=6

2.we calculate the voltage at each point using the equation that relates the distance the voltage and the electric field
V=Ed
where
V=EPD= the electric potential difference
E=electric field magnitude=6.8 × 10^5 N/C.
d=distance
i)
V=Ed=(6.8 × 105 N/C.)(6)=4080000V
ii)
i)
V=Ed=(6.8 × 105 N/C.)(6)=4080000V
iii)
i)
V=Ed=(6.8 × 105 N/C.)(8.48m)=5766400v
Considering Conservation of Momentum, the momentum

before and after must remain the same:
so:
before:
Answer:
Air does, in fact, have weight, and here's a simple way you can prove it. You'll need two identical balloons, a string, and a dowel. Attach the uninflated balloons to either end of the dowel. Attach the string to the center of the dowel and then hang it from something.
Explanation: