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A bullet of mass 0.020kg traveling horizontally at 50m/s is stopped after moving through a distance of 0.10 in a concrete block. What is the average retarding force applied to the bullet by the concrete?

A bullet of mass 0.020kg traveling horizontally at 50m/s is stopped after moving through a distance of 0.10 in a concrete block. What is the average retarding force applied to the bullet by the concrete?

Grade:12

4 Answers

raj
8 Points
7 years ago
v2= u2+ 2as 02= 502+ 2(a)(0.1) a = -12500m/s2 F = ma = 0.020 x (-12500) = -250N The answer is 250N, not -250N. A retarding force is negative in connotation.
RAVI KANT KUMAR
35 Points
7 years ago
impulse=change in momentum t=dist./speed=.1/50=.002 sec impulse=force*time =F*t change in momentum=0-mu=-.02*50=-1 therefore ,F=-1/.0002=5000N
RAVI KANT KUMAR
35 Points
7 years ago
impulse=change in momentum t=dist./speed=.1/50=.002 sec impulse=force*time =F*t change in momentum=0-mu=-.02*50=-1 therefore ,F=-1/.002=500N
EPARI SRAGVEE
19 Points
7 years ago
v2-u2=2as 0=2500+2a[0.1] a= -12500 f=ma =0.020[12500] =-250N

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