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Sagot :
In order to determine the friction force, proceed as follow:
Take into account that total momentum of the system must conserve, then, you have:
m*v = (m + M)v'
where,
m: mass of the bullet = 6.25g = 0.00625kg
v: initial speed of the bullet = 365m/s
M: mass of the crate = 4.50kg
v': speed of both crate and bullet after the impact = ?
Solve the equation above for v', replace tha values of the other parameters and simplify:
[tex]\begin{gathered} v^{\prime}=\frac{mv}{m+M} \\ v^{\prime}=\frac{(0.00625kg)(365\frac{m}{s})}{0.00625\operatorname{kg}+4.50\operatorname{kg}} \\ v^{\prime}\approx0.506\frac{m}{s} \end{gathered}[/tex]Now, consider that the work done by the friction force is given by:
W = Fr*d
where,
Fr: friction force = ?
d: distance = 0.15m
Furthermore, the work done is equal to:
W = 1/2*(m+M)v'^2 that is, the change in kinetic energy is equal to the work
Then, you can equal the previous expressions for W, solve for Fr, replace and simplify:
[tex]\begin{gathered} F_rd=\frac{1}{2}(m+M)v^{\prime}^2 \\ F_r=\frac{1}{2}\frac{(m+M)v^{\prime2}}{d} \\ F_r=\frac{1}{2}\frac{(0.00625kg+4.50kg)(0.506\frac{m}{s})^2}{0.15m} \\ F_r\approx3.846N \end{gathered}[/tex]Now, take into account that the friction force can be written as follow:
Fr = μN = μ(m+M)g
where,
μ: coefficient of kinetic friction between crate and floor
g: gravitational acceleration constant = 9.8m/s^2
Solve the equatio above for μ, replace the values of the other parameters and simplify:
[tex]\begin{gathered} \mu=\frac{F_r}{(m+M)g} \\ \mu=\frac{3.846N}{(0.00625kg+4.50kg)(9.8\frac{m}{s^2})} \\ \mu\approx0.087 \end{gathered}[/tex]Hence, the coefficient of kinetic friction is approximately 0.087
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