A portable CD player uses a current of 6.8 mA at a potential difference of 3 V. (a) How much energy does the p

Question

A portable CD player uses a current of 6.8 mA at a potential difference of 3 V. (a) How much energy does the player use in 41 minutes? (b) Suppose the player has a mass of 0.65 kg. How many minutes should the player operate on so that the energy can lift the CD player through a height of 1m?
(a) Energy consumed in 41 minutes is J.
(b) The time for the CD player is minutes.

I am extremely bad with physics so along with an explanation, and answer would be much appreciated! Thanks!

Answer 1

(a) (3 V)(6.8 mA)(1 A/1,000 mA)(41 min)(60 sec/1 min) = 50.184 watt-sec ≈ 50 J

(b) (0.65 kg)(9.81 m/s^2)(1 m) / ((3 V)(6.8 mA)(1 A/1,000 mA)(60 sec/1 min)) ≈ 5.2 min.

Answer 2

A)Energy = Charge x Voltage = Current x Time x Voltage
(E = Q x V = I x T x V )
So, E = 3 x 6.8exp-03 x (41 x 60) = 50.184J

B)Gravitational potentional energy = mass x gravity x height
GPE = MGH
So, 0.65 x 9.8 x 1 = 6.37J is the energy required.
The player uses 50.184J in 41 minutes.
50.184 / 6.37 = 7.878
60 x 40 (total seconds in 41 minutes) = 2460
2460/7.878 = 312.26 seconds.
That in minutes is 6 minutes and12 seconds.
I couldn't think how to make that last calculation easier, i know there's a way but it's not in my head at the minute!
Of course I might have got something wrong somewhere along the line.