1) A 11 kg crate is moved from the ground up a loading platform 6.3 m long and 1

Question

1) A 11 kg crate is moved from the ground up a loading platform 6.3 m long and 1.8 m high. Assume the coefficient of kinetic friction is 0.23. What is the actual mechanical advantage and incline plane efficiency?

Actual MA = XXX ***prior answer provided by Chegg of actual MA = 0.98 was incorrect for this question***

Incline Efficiency = %

2) During the operation of a 320 hp engine, energy is lost to friction at a rate of 140 hp. What is the output power and the engine efficiency?

Output Power = XXX horsepower

Efficiency = XXX %

Explanation / Answer

Givens: m = 11kg; l = 6.3m and h = 1.8m, coefficient of kineic friction: 0.23

Mechanical Advantage: ?

For the incline, the angle of incline , theta = arcsin(1.8/6.3) --> theta = 18.45 degree

Now , force needed to move up the incline:

F = mg* sin(theta) + uk * mg * cos(theta)

F = 11 * 9.8 *(sin(18.45) + 0.23 * cos(18.45)) -->

F = 54.7 N

mechanical advantage = mg/F

mechanical advantage = 9 * 9.8/54.7

mechanical advantage = 2 --> the actual mechanical advantage of the ramp is 1.613

Now , for efficiency

efficiency = energy gained by block/total energy used

efficiency = 11 * 9.8 * 1.8 /( 11* 9.8 * 1.8 + 0.23 * 11 * 9.8 * cos(24.2) * 6.3)

efficiency = 0.58 --> the efficiency of the plane is 58%

**********************************************************************************************************************

2) Poutput = 320hp -140hp = 180hp

Efficiency = Poutput / Pinput ---> Efficiency = 180hp / 320hp ---> Efficiency = 0.5625 ---> Efficiency = 56.25%

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