a.)
m1v1+m2v2=m3v3
Let:
m1=1300g=1.3kg
v1=0m/s
m2=39g=0.039kg
v2=290m/s
m3=1.339kg *** addition of m1 and m2
and
v3=???
Solve for v3:
(1.3kg+0)+(0.039kg+290m/s)=(1.339kg)v3
v3=[(1.3kg+0)+(0.039kg+290m/s)]/1.339kg
v3=1.3+11.31/1.339
v3=9.41m/s
b.)
Fnet=F(block)-F(friction)
Let
Fnet=m3*a
F(block)=m3*g=(1.339kg*9.8m/s^2)
**g=gravity=9.8m/s^2
and
F(friction)=m3*g*(0.74)
Fnet=13.12N-9.7N
Fnet=3.42N
3.42N=1.339kg*a
a=3.42N/1.339kg
a=2.55m/s^2
c.)
In order for the block to stop, the work done by kinetic friction must equal 0. That is, the force of friction does takes away all of the kinetic energy of the system.
0=Kinetic Energy-Work done by Friction
0=1/2mv^2-F*d
F*d=1/2mv^2
d=(1/2mv^2)/F
let
m=m3=1.339kg **From a
v=v3=9.41m/s ***From a
and
F=9.7N *** From b
Solve for d:
d=(0.5(1.339kg*9.41m/s^2)^2)/9.7N
d=6.12m
*** Hopefully, someone checks this.
A block of wood with a mass of 1300g is sitting on a horizontal tabletop. The coefficient of kinetic friction between the block and table is µk=0.74.A rifle bullet with a mass of 39g is fired horizontally into the block at a sped of 290m/s and stops inside the block.
a-what is the velocity of the bullet/block system after impact?
b-what is the acceleration/deceleration experienced by the bullet/block system due to the friction?
c-how far does the block with the bullet embedded in it slide before coming to a stop?
1 answer
1 answer