Answer:
Change in potential energy of the block-spring-Earth
system between Figure 1 and Figure 2 = 1 Nm.
Explanation:
Here, spring constant, k = 50 N/m.
given block comes down eventually 0.2 m below.
here, g = 10 m/s.
let block be at a height h above the ground in figure 1.
⇒In figure 2,
potential energy of the block-spring-Earth
system = m×g×(h - 0.2) + 1/2× k × x². where, x = change in spring length.
⇒ Change in potential energy of the block-spring-Earth
system between Figure 1 and Figure 2 = (m×g×(h - 0.2)) - (1/2× k × x²)
= (1×10×0.2) - (1/2×50×0.2×0.2) = 1 Nm.
Fortunately, 'force' is a vector. So if you know the strength and direction
of each force, you can easily addum up and find the 'resultant' (net) force.
When we talk in vectors, one newton forward is the negative of
one newton backward. Hold that thought, while I slog through
the complete solution of the problem.
(100 N forward) plus (50 N backward)
= (100 N forward) minus (50 N forward)
= 50 N forward .
That's it.
Is there any part of the solution that's not clear ?
1) The grapefruit is in free fall, so it moves by uniformly accelerated motion, with constant acceleration
. Calling h its height at t=0, the height at time t is given by
We are told thatn when
the grapefruit hits the ground, so h(0.75 s)=0. If we substitute these data into the equation, we can find the initial height h of the grapefruit:
2) The speed of the grapefruit at time t is given by
where
is the initial speed of the grapefruit. Substituting t=0.75 s, we find the speed when the grapefruit hits the ground:
The objects will remain at rest if net force acting on it is zero if their magnitude is same and they are acting in opposite direction then according to Newton's 2nd law the net force acting on the system is zero. Since the net force on the system is zero, the object will remain at rest.
Answer:
3494444444.44444 J
-87077491.39453 J
Explanation:
M = Mass of Earth = 
G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²
R = Radius of Earth = 
h = Altitude = 
m = Mass of satellite = 629 kg
v = Velocity of spacecraft = 
The kinetic energy is given by
The spacecraft's kinetic energy relative to the earth is 3494444444.44444 J
Potential energy is given by
The potential energy of the earth-spacecraft system is -87077491.39453 J
