Question

a sack rice weighing 500 newton's is pulled 10 metres is along a level floor with a force of 50 newton's .what is the work done on the sack of rice ?​

Answer 1

Answer:

Correct option is

A

1.2×10

5

J

Given,

Initial velocity, u=72 km/h

u=

3600

72×1000

m/s=20 m/s

Final velocity, v=0 m/s

Displacement s=60 m

Using Newton's third equation of motion,

2as=v

2

−u

2

⟹a=

2s

v

2

−u

2

⟹a=

2×60

0

2

−20

2

⟹a=

2×60

0

2

−20

2

⟹a=

120

400

a=−

3

10

m/s

2

Magnitude of acceleration, a=

3

−10

m/s

2

Force F=ma=600×(10/3)

F=2000 N

Work done, W=F×S

W=2000×60

W=120000J

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Answer 2

Answer:

workdone = force × distance

workdone = 50N × 10m

workdone = 500Nm

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