Physics, 12.08.2020 09:01 jdkrisdaimcc11
A worker wants to load a 12 kg crate into a truck by sliding the crate up a straight ramp which is 2.5 m long and which makes an angle of 30 degrees with the horizontal. The worker believes that he can get the crate to the very top of the ramp by launching it at 5 m/s at the bottom and letting go. But friction is not neglible; the crate slides 1.6 m upthe ramp, stops, and slides back down.
Required:
a. Assuming that the friction force actingon the crate is constant, find its magnitude.
b. How fast is teh crate moving when it reachesthe bottom of the ramp?
Answers: 1
, 
- Initial and final gravitational potential energy, measured in joules. 
, 
- Initial and final translational kinetic energy, measured in joules.
- Work losses due to friction, measured in joules.
- Mass of the crate, measured in kilograms.
- Gravitational acceleration, measured in meters per square second.
, 
- Initial and final height of the crate, measured in meters.
, 
- Initial and final speeds of the crate, measured in meters per second.
- Kinetic coefficient of friction, dimensionless.
- Ramp inclination, measured in sexagesimal degrees.
![\mu_{k} = \frac{1}{\cos \theta} \cdot \left[y_{1}-y_{2}+\frac{1}{2\cdot g}\cdot (v_{1}^{2}-v_{2}^{2}) \right]](/tpl/images/0720/3363/cf055.png)
, 
, 
, 
and 
, the coefficient of friction is:![\mu_{k} = \frac{1}{\cos 30^{\circ}}\cdot \left\{0\,m-0.8\,m+\frac{1}{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)}\cdot \left[\left(5\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right] \right\}](/tpl/images/0720/3363/58819.png)
, 
, 
, 
, the speed of the crate at the bottom of the ramp is:![v_{2}=\sqrt{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot [0.8\,m-0\,m-(0.548)\cdot \cos 30^{\circ}]+\left(0\,\frac{m}{s} \right)^{2}}](/tpl/images/0720/3363/4f2ef.png)
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