Magnitude of the Frictional Force decelerating a bullet that hits a block

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Homework Statement:
A 7.80 g bullet is initially moving at 500 m/s just before it penetrates a block of solid rubber to a depth of 4.50 cm.
(a)What is the magnitude of the average frictional force (in N) that is exerted on the bullet while it is moving through the block of solid rubber? Use work and energy considerations to obtain your answer.
(b)Assuming the frictional force is constant, how much time (in s) elapses between the moment the bullet enters the block of solid rubber and the moment it stops moving?
Relevant Equations:
.
A 7.80 g bullet is initially moving at 500 m/s just before it penetrates a block of solid rubber to a depth of 4.50 cm.
(a)What is the magnitude of the average frictional force (in N) that is exerted on the bullet while it is moving through the block of solid rubber? Use work and energy considerations to obtain your answer.
(b)Assuming the frictional force is constant, how much time (in s) elapses between the moment the bullet enters the block of solid rubber and the moment it stops moving?
 

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  • #2
BvU
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Hello @shin ,
:welcome: !​

Here at PF things don't go as you seem to expect: dump an exercise and get the answer.
Please read the PF guidelines and post your attempt.

Oh, and: does the rubber block stay in place or can it e.g. fly off because it's hanging from a wire ?
 
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haruspex
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Homework Statement:: A 7.80 g bullet is initially moving at 500 m/s just before it penetrates a block of solid rubber to a depth of 4.50 cm.
(a)What is the magnitude of the average frictional force (in N) that is exerted on the bullet while it is moving through the block of solid rubber? Use work and energy considerations to obtain your answer.
(b)Assuming the frictional force is constant, how much time (in s) elapses between the moment the bullet enters the block of solid rubber and the moment it stops moving?
This is a distressingly frequent blunder by question setters who ought to know better.
Average force is ##F_{avg}=\frac{\Delta p}{\Delta t}##, the change in momentum divided by elapsed time. In vectors, ##\vec F_{avg}=\frac{\vec{\Delta p}}{\Delta t}##.
Note that this is consistent with acceleration and velocity. Cancelling mass out we get:
##a_{avg}=\frac{\Delta v}{\Delta t}##.
In general, this is not the same as ##\frac{\Delta E}{\Delta s}##, where E is energy and Δs is displacement in the direction of the force. They will be the same if the force is constant, but it cannot be written as a vector equation because energy is a scalar and you cannot divide by a vector.

Hence the correct wording of the question may be:
Assume the frictional force is constant during the penetration.
a) What is its magnitude?
b) How long does the penetration take?
 
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