Answer:
the energy of the spring at the start is 400 J.
Explanation:
Given;
mass of the box, m = 8.0 kg
final speed of the box, v = 10 m/s
Apply the principle of conservation of energy to determine the energy of the spring at the start;
Final Kinetic energy of the box = initial elastic potential energy of the spring
K.E = Ux
¹/₂mv² = Ux
¹/₂ x 8 x 10² = Ux
400 J = Ux
Therefore, the energy of the spring at the start is 400 J.
Answer:
d) I and III only.
Explanation:
Let be 
and 
the masses of the two laboratory carts and let suppose that 
. The expressions for each kinetic energy are, respectively:
and 
.
After some algebraic manipulation, the following relation is constructed:
Since 
, then 
. That is to say, 
.
The expressions for each linear momentum are, respectively:
and 
Since , then 
. Which proves that statement I is true.
According to the Impulse Theorem, the impulse needed by cart I is greater than impulse needed by cart II, which proves that statement II is false.
According to the Work-Energy Theorem, both carts need the same amount of work to stop them. Which proves that statement III is true.
Answer:
Increase
Explanation:
because it's increasing up.
Answer:
The block+bullet system moves 4 m before being stopped by the frictional force.
Explanation:
Using the law of conservation of llinear momentum and the work energy theorem, we can obtain this.
According to Newton's second law of motion
Momentum before collision = Momentum after collision
Momentum before collision = (0.02×400) + 0 (stationary block)
Momentum before collision = 8 kgm/s
Momentum after collision = (2+0.02)v
8 = 2.02v
v = 3.96 m/s.
According to the work-energy theorem,
The kinetic energy of the block+bullet system = work done by Friction to stop the motion of the block+bullet system
Kinetic energy = (1/2)(2.02)(3.96²) = 15.84 J
Work done by the frictional force = F × (distance moved by the force)
F = μmg = 0.2(2.02)(9.8) = 3.96 N
3.96d = 15.84
d = (15.84/3.96) = 4 m
when you drop a ball and let it fall to the ground
