Question

Juan lifts a 7.5 kilogram rock 1.75 meters in 5.0 seconds. How much work is done,

and what power is used?

Answer 1

The work done by Juan in lifting an object 1.75 m high is 128.625 J.

The power used in 5.0 s is 25.725 W.

Step-by-step explanation:

Given:

Mass of the object is,
`m=7.5\ kg`

Displacement of the object is,
`d=1.75\ m`

Time taken for the displacement caused is,
`t=5.0\ s`

Acceleration of the object is due to gravity,
`g=9.8\ m/s^2`

The work done by Juan to lift the object will be equal to the force applied by Juan and displacement caused due to the force in the direction of the force applied.

Here, the force applied by Juan will be equal to gravitational force but in the upward direction.

Force applied is given as:

`F=mg`

So, work done is given as:

`W=F\cdot h=mgh\\W=7.5* 9.8* 1.75\\W=128.625\ J`

Therefore, the work done by Juan in lifting an object 1.75 m high is 128.625 J.

Now, power used is given as:

`P=(W)/(t)\\P=(128.625)/(5.0)=25.725\ W`

Therefore, the power used in 5.0 s is 25.725 W.

Answer 2

The work done W = = 128.63 J

The power, p = 25.7 watts

Step-by-step explanation:

Given,

The mass of the rock, m = 7.5 kg

The displacement of the rock, s = 1.75 m

The time period of work, t = 5 s

The force acting on the body,

F = m x g

= 7.5 kg x 9.8 m/s²

= 73.5 N

The work done is given by the relation,

W = F x S

Substituting the given values,

W = 73.5 x 1.75

= 128.63 J

The power is defined as the rate of doing work

P = W / t

= 128.63 / 5

= 25.7 watts

Hence, the power, p = 25.7 watts