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. The crate is speeding up at 0.33 m/s2. The static and kinetic coefficients of friction are 0.42 and 0.30, respectively. What is the work done by you on the crate? 2.) You drive your 1,309-kg car at a constant speed of 21 mph down a ramp that has a grade of 12 %. Use conservation of energy to find the work done by the frictional force acting from the road on the tires. Define positive direction in the forward direction. If you could please show me how this is done I would greatly appreciate it. Thank you :)

Explanation / Answer

The crate is moving, so you only need to use the kinetic coefficient of friction: Force of friction = Kinetic coefficient*mass*gravitational acceleration Force of friction = 0.30*150kg*9.81m/s^2 = 441.45 Newtons Work = Force*Distance = 441.45N*4.5m = 1986.525 Joules Total Force = Force applied by you - Force of friction Total Force = mass*acceleration = 150kg*0.38m/s^2 = 57 Newtons Force applied by you = 441.45 Newtons + 57 Newtons = 490.95 Newtons Work done by you = Force applied by you*Distance = (490.95N)(4.5m) = 2209.275 Joules

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