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A Sliding Crate of Fruit?
A crate of fruit with a mass of 37.5 kg and a specific heat capacity of 3650 J/(kg *K) slides 7.70 m down a ramp inclined at an angle of 36.2 degrees below the horizontal.
A) If the crate was at rest at the top of the incline and has a speed of 2.50 m/s at the bottom, how much work W_f was done on the crate by friction?
2 Answers
- JicotilloLv 61 decade agoFavourite answer
Work done by friction can be found as the difference of potential energy and kinetic energy.
Potential energy change: mgh = mgs sin 36.2°.
Kinetic energy change: ½ mv².
Work done by friction (on crate): mgs sin 36.2° − ½ mv²
W = 37.5(9.8 × 7.7 sin 36.2° − ½ × 2.5²) = 1 554 J.
For a frictionless incline surface, potential energy change = kinetic energy change, along with null friction work. In the present case, however, kinetic energy is only about 7% of potential energy: almost 93% of energy is converted to heat.
