The final speed of the crate is 6.3 m/s
Explanation:
Please note that there is a typo in the text of the question: the coefficient of sliding friction is 0.15, not 9.15.
First of all, we need to find the force of friction acting on the crate. This is given by:
![F_f = \mu mg](https://tex.z-dn.net/?f=F_f%20%3D%20%5Cmu%20mg)
where:
is the coefficient of friction
m = 50 kg is the mass of the crate
is the acceleration of gravity
Substituting,
![F_f = (0.15)(50)(9.8)=73.5 N](https://tex.z-dn.net/?f=F_f%20%3D%20%280.15%29%2850%29%289.8%29%3D73.5%20N)
Now we can find the acceleration of the crate by using Newton's second law:
![F-F_f = ma](https://tex.z-dn.net/?f=F-F_f%20%3D%20ma)
where
F = 90.0 N is the force applied forward on the crate
is the force of friction, acting backward
a is the acceleration of the crate
Solving for a,
in the forward direction.
Finally, we can find the final speed of the crate by using suvat equations:
![v^2-u^2 = 2as](https://tex.z-dn.net/?f=v%5E2-u%5E2%20%3D%202as)
where:
v is the final speed
u = 0 is the initial speed
is the acceleration
s = 60 m is the distance covered
Solving for v,
![v=\sqrt{u^2+2as}=\sqrt{0+2(0.33)(60)}=6.3 m/s](https://tex.z-dn.net/?f=v%3D%5Csqrt%7Bu%5E2%2B2as%7D%3D%5Csqrt%7B0%2B2%280.33%29%2860%29%7D%3D6.3%20m%2Fs)
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