# A small electric motor with a power rating of 3.5 W is mounted on top of a desk. It is used to lift a 1.2kg mass up from the floor to the desk top in a time of 2.8 seconds. How high is the desk? If the 1.2 kg mass is replaced by a larger 2.0kg mass, how much time will it take for the motor to lift the new mass from the floor to the desk top? To solve for the height of the desk, apply the formula that relates power to work which is:

`P = W/ (Delta t)`

where

P is power,

W is the work done, and

`Delta t` is the time interval.

Plugging in the given values P=3.5W and `Delta t ` = 2.8s, the formula becomes:

`3.5 = W/2.8`

Isolating the W, it yields:

`3.5*2.8=W/2.8*2.8`

`9.8=W`

So, the work done in lifting the object from the floor to the desk is 9.8J.

Take note that work is calculated by multiplying the force  by the amount of movement.

`W = F * Delta x`

where

F is the force and

`Delta x` is the displacement.

In this problem, the only force present in lifting the object is its weight.

`W = mg * Delta x`

Plugging in the values W=9.8J, m=1.2kg and g=9.81 m/s^2, the formula becomes:

`9.8=1.2*9.81*Delta x`

Isolating Delta x results to:

`9.8=11.772*Delta x`

`9.8/11.772=(11.772*Delta x)/11.772`

`0.8324 = Delta x`

Therefore, the desk is 0.8324 meters high.

To solve for the time it took to lift the new object that has a mass of 2kg, the same formulas are to be used. However, the order of the steps will be different.

Here, the work done in lifting the object should be computed first. Plugging in the values m=2kg, g=9.81 m/s^2 and `Delta x` =0.8324m, the work done is:

`W=F*Delta x=mg*Delta x=2*9.81*0.8324=16.3317`

Then, plug in P=3.5W and W=16.3317J to the formula of power.

`P=W/(Delta t)`

`3.5=16.3317/(Delta t)`

Isolating the `Delta t`, the formula becomes:

`3.5*Delta t= 16.3317/(Delta t)*Delta t`

`3.5*Delta t = 16.3317`

`(Delta t)/3.5=16.3317/3.5`

`Delta t=4.67`

Therefore, it took the motor 4.67 seconds to the lift the new object from the floor to the desk.

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