Question

A crate of fruit with mass 35.0kg and specific heat capacity 3650 J/kg K slides down a ramp inclined at 36.8 degrees below the horizontal. The ramp is 8.00m long.a) if the crate was at rest at the top of the incline and has a speed of 2.50m/s at the bottom, how much work was done on the crate by friction?b) if an amount of heat equal to the magnitude of the work done by friction goes into the crate of fruit and the fruit reaches a uniform final temperature, what is its temperature change?

Answer 1

Answer:

a)    W = 1120 J , b)   T = 36.809 C

Explanation:

a) To find work, use the relationship between work and kinetic energy

            W = ΔK

Since the box starts from rest, its initial speed is zero.

             W = ½ m v² - 0

Let's calculate

           W = ½ 35 8²

          W = 1120 J

b) this work is transformed into heat in the box, let's look for the increase in temperature

          Q = W

          Q = m $c_{e}$ (T - T₀)

        T = T₀ + Q / m c_{e}

The specific heat in degrees kelvin and centigrade has the same numerical value since the variation of the two units is the same

       

Let's calculate

          T = 36.8 + 1120 / (35 3650)

          T = 36.8 + 0.0088

          T = 36.809 C

Answer 2

A. The work done on the crate by friction is 109.375 J

B. The temperature change of the fruit is 8.56×10¯⁴ K

A. Determination of the work done

Mass (m) = 35 Kg

Initial velocity (u) = 0 m/s

Final velocity (v) = 2.5 m/s

Work done (Wd) =?

Work done = change in energy

Change in energy = ½m(v – u)²

= ½ × 35 × (2.5 – 0)²

= 17.5 × 2.5²

Work done = 109.375 J

B. Determination of the change in the temperature of the fruit

Work = 109.375 J

Heat (Q) = work = 109.375 J

Mass (M) = 35 Kg

Specific heat capacity (C) = 3650 J/kgK

Change in temperature (ΔT) =?

Q = MCΔT

109.375 = 35 × 3650 × ΔT

109.375 = 127750 × ΔT

Divide both side by 127750

ΔT = 109.375 / 127750

ΔT = 8.56×10¯⁴ K

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