Answer:
Given force=10lb
L1=4in converting to feet
But 0.08333ft= 1 inch
Then 4 inch is 0.3332
6inch is 0.49998
But hookes law states
F=Kx where F is force,K is the force constant ,X
K=F/X=10/0.3333=30N/m
Integrating this
Integral of 30x with limit 0.333 to 0.5
F=30x^2/2=15x^2substing the limit
F=(15(0.5^2-0.33^2)=2.08ft-lb
Explanation:
Kinetic energy = (1/2)*mass*velocity^2
KE = (1/2)mv^2
KE = (1/2)(478)(15)^2
KE = 53775J
Answer:
speed of the bullet before it hit the block is 200 m/s
Explanation:
given data
mass of block m1 = 1.2 kg
mass of bullet m2 = 50 gram = 0.05 kg
combine speed V= 8.0 m/s
to find out
speed of the bullet before it hit the block
solution
we will apply here conservation of momentum that is
m1 × v1 + m2 × v2 = M × V .............1
here m1 is mass of block and m2 is mass of bullet and v1 is initial speed of block i.e 0 and v2 is initial speed of bullet and M is combine mass of block and bullet and V is combine speed of block and bullet
put all value in equation 1
m1 × v1 + m2 × v2 = M × V
1.2 × 0 + 0.05 × v2 = ( 1.2 + 0.05 ) × 8
solve it we get
v2 = 200 m/s
so speed of the bullet before it hit the block is 200 m/s
Gravitational acceleration is approx 9.8 m/s
Time is 7s
a=9.8 m/s
t=7s
a = d/t^2
therefore:
d = a * t^2
d = 9.8 * 7^2
d = 9.8 * 49
d = 480.2 [m]
