80,860 views
6 votes
6 votes
How long will it take a 50w motor to lift a 20 kg object 5 m?

User Andreas Gelever
by
3.1k points

2 Answers

11 votes
11 votes

Answer:

50+20=70

70÷5=14

Step-by-step explanation:

50×20=1000

1000÷5=200

User KosiB
by
2.9k points
16 votes
16 votes

Answer:

Approximately
20\; {\rm s}, assuming no energy loss and that
g = 10\; {\rm N \cdot kg^(-1)}.

Step-by-step explanation:

If the gravitational field strength is constantly
g, lifting an object of mass
m by a height of
\Delta h would increase the gravitational potential energy (
\text{GPE}) of that object by
m\, g\, \Delta h.

In this question,
m = 20\; {\rm kg} and
\Delta h = 5\; {\rm m}. Assume that
g = 10\; {\rm N \cdot kg^(-1)}. Hence, the
\text{GPE} that this object would eventual gain would be:


\begin{aligned}m\, g\, \Delta h &= 20\; {\rm kg} * 10\; {\rm N \cdot kg^(-1)} * 5\; {\rm m} \\ &= 1000\; {\rm N \cdot m} \\ &= 1000\; {\rm J}\end{aligned}.

Thus, this motor would need to do (at least)
1000\; {\rm J} of work to lift this object up by
5\; {\rm m}. The power rating of this motor,
P = 50\; {\rm W} = 50\; {\rm J \cdot s^(-1)}, means that (assuming no energy loss) this motor could do
50\; {\rm J} of work every second.

The time it would take for this motor to do
1000\; {\rm J} of work under these assumptions would be:


\begin{aligned}t &= \frac{\text{work}}{\text{power}} \\ &= \frac{1000\; {\rm J}}{50\; {\rm J \cdot s^(-1)}} \\ &= 20\; {\rm s}\end{aligned}.

User Bunmi
by
2.9k points