Answer:
.
Explanation:
The efficiency of a machine is the percentage of energy input that was turned into useful energy.
The power rating of this lamp is 
(same as 
,) meaning that 
of energy is supplied to this lamp every second.
The question states that 
out of that of energy input would be turned into heat, which is not useful energy output in this scenario. Assuming that all other forms of energy loss is negligible. The rest of the 
of energy supplied to this lamp would be turned into useful energy output.
Thus, every second, this lamp would receive of energy input and would outputs 
of useful work. The efficiency of this lamp would be:
.
Explanation:
Start with what you know and list your knowns and unknowns
F = ma
F= 3N
m = 6kg
a =?
3N = 6kg x a
solve for a
3N / 6kg = a
