A crate of fruit with a mass of 40.0kg and a specific heat capacity of 3500J/kgK

Question

A crate of fruit with a mass of 40.0kg and a specific heat capacity of 3500J/kgK slides 7.50m down a ramp inclined at an angle of 40.0degrees below the horizontal. If the crate was at rest at the top of the incline and has a speed of 2.90m/s at the bottom, how much work was done on the crate by friction? If an amount of heat equal to the magnitude of the work done by friction is absorbed by the crate of fruit and the fruit reaches a uniform final temperature, what is its temperature change delta T?

Explanation / Answer

work done by friction= - 0.83*40*9.8*0.64*7.5=-1561.78j ms(t2-t1)=0.5m*v^2+(-1561.78).....change in temperature=0.009953degrees centigrade

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.