A Sliding Crate of Fruit A crate of fruit with a mass of 35.0kg and a specific h

Question

A Sliding Crate of Fruit A crate of fruit with a mass of 35.0kg and a specific heat capacity of 3500J/(kg . K) slides 8.70m down a ramp inclined at an angle of 37.7degrees below the horizontal. Part A If the crate was at rest at the top of the incline and has a speed of 3.00110 at the bottom, how much work Wf was done on the crate by friction? Use 9.81m/s2 for the acceleration due to gravity and express your answer in joules. Wf = [ ] Part B This question will be shown after you complete previous question(s).

Explanation / Answer

Work done by gravitational force is wg = Fg*S*sin(0) = m*g*S = 35*9.81*8.7 = 2987.145 J

Work done by Frictional force Wf = ?

Work done by net force = Wg+Wf

According to Work Energy theorem

Wg+Wf = 0.5*m*v^2 =0.5*35*3^2 = 157.5

Wf = 157.5 - 2987.145 = -2829.645 J

negative sign represents loss in energy

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