Question

a constant force acting on a 20 kg mass change its velocity from 30 km per hour to 60 km per hour calculate work done by force​

Answer 1

Answer:

Given : Initial velocity u=5 m/s

Final velocity v=2 m/s

The work done is equal to the change in the kinetic energy of an object.

W=

2

1

m(v

2

−u

2

)

W=

2

1

×20×(2

2

−5

2

)

⟹W=−210 J

Explanation:

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Answer 2

Answer:

the answer is 1215.28 J

Explanation:

since a constant force is acting on 20kg mass

we can apply here work energy theorem,

According to Work Energy theorem

Work done = ∆ KE

where '∆' represents change

KE is Kinetic energy. Work done is change in Kinetic energy.

mass given = 20kg

We need to convert 30km/hr and 60km/hr into m/s to get the work done in joules.

therefore 30km/hr in m/s

1km = 1000m

1 hour = 3600 s

$30 \times \frac{1000}{3600} = 30 \times \frac{5}{18} = \frac{25}{2} {ms}^{ - 1}$

$60 \times \frac{1000}{3600} = 60 \times \frac{5}{18} = \frac{50}{3} {ms}^{ - 1}$

therefore putting in work energy theorem we get,

$W=∆ KE \\ = \frac{1}{2} \: m {(vfinal)}^{2} - \frac{1}{2} \: m {(vinitial)}^{2} \\ = \frac{1}{2} m({(vfinal)}^{2} - {(v \: initial)}^{2}) \\ = \frac{1}{2} m( { (\frac{50}{3} })^{2} - { (\frac{25}{2} })^{2} ) \\ = \frac{1}{2} m( \frac{2500}{9} - \frac{625}{4} ) \\ = \frac{1}{2} m( \frac{10000 - 5625}{36} ) \\ = \frac{1}{2} m( \frac{4375}{36} ) \\ = \frac{1}{2} m(121.528) \\ = \frac{1}{2} \times 20 \times 121.528 \\ = 1215.28J$

hope it helps.