A bullet with a mass of 0.02 kg is fired horizontally into a block of wood hanging on a string. The bullet sricks in the wood an
d causes it to swing upward to a height of 0.2 m. If the mass of the wood block is 2 kg, what is the initial speed of the bullet?
1 answer:
Answer:
u= 20.09 m/s
Explanation:
Given that
m = 0.02 kg
M= 2 kg
h= 0.2 m
Lets take initial speed of bullet = u m/s
The final speed of the system will be zero.
From energy conservation
1/2 m u²+ 0 = 0+ (m+M) g h
m u²=2 (m+M) g h
By putting the values
0.02 x u² = 2 (0.02+2) x 10 x 0.2 ( take g=10 m/s²)
u= 20.09 m/s
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Answer:
The time taken to stop the box equals 1.33 seconds.
Explanation:
Since frictional force always acts opposite to the motion of the box we can find the acceleration that the force produces using newton's second law of motion as shown below:
Given mass of box = 5.0 kg
Frictional force = 30 N
thus
Now to find the time that the box requires to stop can be calculated by first equation of kinematics
The box will stop when it's final velocity becomes zero
Here acceleration is taken as negative since it opposes the motion of the box since frictional force always opposes motion.