1. You are pushing a 150-kg wooden crate in a straight line a distance of 4.5 m

Question

1. You are pushing a 150-kg wooden crate in a straight line a distance of 4.5 m across a wooden floor at constant speed. The static and kinetic coefficients of friction are 0.42 and 0.30, respectively. What is the work done by friction on the crate?

  ( I used the formula m*.30*g*d and i got it wrong)

2. Converting the energy of one food calorie into mechanical energy, starting from rest, to what speed could you accelerate a 194-lbs person?

    (I need the formula for this problem, everyone writes the answer but not the formula)

3. You are pushing a 150-kg wooden crate in a straight line a distance of 4.5 m across a wooden floor. The crate is speeding up at 0.22 m/s2. The static and kinetic coefficients of friction are 0.42 and 0.30, respectively. What is the work done by friction on the crate?

Explanation / Answer

1) Ff = -m*g*mue_s*d = -150*9.8*4.5*0.3 = -1984.5 J

2)energy given to the person = 0.5*m*v^2

3)
Ff = -m*g*mue_s*d = -150*9.8*4.5*0.3 = -1984.5 J

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