Question

Shoveling snow can be extremely taxing since the arms have such a low efficiency in this activity. Suppose a person shoveling a sidewalk metabolizes food at the rate of 700 W. (The efficiency of a person shoveling is 3%.) (a) What is her useful power output (in W)? W (b) How long (in s) will it take her to lift 2,700 kg of snow 1.20 m? (This could be the amount of heavy snow on 60.0 ft of sidewalk.)

Answer 1

A) We know that efficiency in a system is defined by the equation

`\eta = (P_(out))/(P_(in))`

Where P is the Power in a system. Then,

`P_(out) = P_(in)* \eta`

`P_(out) = 700*3\%`

`P_(out) = 21W`

Therefore the output power is 21W

B) Work is defined by,

`W=F*d`

`W=mg*d=2700(9.8)*1.2 = 31.75kJ`

To calculate the power we know that is the rate of work in a determined time, then

`P=(W)/(t)`

Re-arrange for t,

`t=(W)/(P) = (31.75*10^3)/(21)=1511s`

c) Finally we have that the amount of heat wasted depends on the power, then,

`Q= (700-21)*1511`

`Q=1.02*10^6J`