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Answer:
The final speed of the crate is 12.07 m/s.
Explanation:
For the first 10.0 meters, the only force acting on the crate is 225 N, so we can calculate the acceleration as follows:
[tex] F = ma [/tex]
[tex] a = \frac{F}{m} = \frac{225 N}{51.0 kg} = 4.41 m/s^{2} [/tex]
Now, we can calculate the final speed of the crate at the end of 10.0 m:
[tex] v_{f}^{2} = v_{0}^{2} + 2ad_{1} [/tex]
[tex] v_{f} = \sqrt{0 + 2*4.41 m/s^{2}*10.0 m} = 9.39 m/s [/tex]
For the next 10.5 meters we have frictional force:
[tex] F - F_{\mu} = ma [/tex]
[tex] F - \mu mg = ma [/tex]
So, the acceleration is:
[tex] a = \frac{F - \mu mg}{m} = \frac{225 N - 0.17*51.0 kg*9.81 m/s^{2}}{51.0 kg} = 2.74 m/s^{2} [/tex]
The final speed of the crate at the end of 10.0 m will be the initial speed of the following 10.5 meters, so:
[tex] v_{f}^{2} = v_{0}^{2} + 2ad_{2} [/tex]
[tex] v_{f} = \sqrt{(9.39 m/s)^{2} + 2*2.74 m/s^{2}*10.5 m} = 12.07 m/s [/tex]
Therefore, the final speed of the crate after being pulled these 20.5 meters is 12.07 m/s.
I hope it helps you!