First method
initial distance = 16m
final distance= 43 m
total distance covered= final -initial
=43m -16m
=27m
Second method
Si= 16m
Sf =43 m
t= 12 s
first we will find V
V = (Sf-Si)/ t
V =( 43- 16)/ 12
V = 27/12 ⇒ V= 9/4
V= distance / time
distance= V×time
distance = (9/4) ×12
distance =27
A) work = force * distance
mass is not a force, weight is, so we have to find the weight of the block.
Weight = mg
Weight = (220kg)(9.8)
Weight = 2156N
Work = 2156N * 3.10m
work = 6683.6J
b) Since he is holding the weights, it's not moving, therefore, he doesn't do any work
c) The answer is still the same amount of work when he lifted them.
d) The answer is no since when he let go the weight, he doesn't apply any force to the weight.
e) P = work/time
P = 6683.6J / 2.1s
P = 3182.67 watts
To find the rate of deceleration, subtract the initial velocity from the final velocity.
The block with the bullet lodged in the block is now travelling at 2.133 m/s.
<h3>What is momentum conservation principle?</h3>
When there is no external force acting on the system, the momentum remains conserved.
For inelastic collision, after collision both objects travel with common speed.
m1u1 + m2u2 =(m1 +m2)v
Substitute initial velocity of bullet u1 =320 m/s , initial velocity of block u2 =0, mass of bullet m1 = 0.1 kg and mass of block m2 = 14.9 kg.
Solve for the final velocity of bullet,
0.1 x 320 + 14.9 x 0 = (0.1 +14.9) x v
v = 2.133 m/s
Thus, the block with the bullet lodged in block now travelling at 2.133 m/s.
Learn more about momentum conservation principle.
brainly.com/question/14033058
#SPJ1
