Calculate the work done in lifting a 300Kg weight to a height of 10m with the accelsration 0.5 ms-2. Take g=10ms-1
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F=ma
=300×0.5
=150N
S=10m
W=fs
=150×10
1500j
=300×0.5
=150N
S=10m
W=fs
=150×10
1500j
Answered by
2
Answer:
Explanation:
It is given that,
Weight = 300 N
height ( h ) = 10 m
acceleration ( a ) = 0.5 m/s²
Value of ' g ' = 10 m/s²
Now,
In order to find the work done by the body, we've to find it's mass and total acceleration acting on the body. And then finally, substitute the values in the formula :-
Workdone = mgh
∵ weight = mg
→ 300 = m × 10
→ m = 300/10
∴ mass = 30 kg
TOTAL ACCELERATION ACTING ON THE BODY :
→ 0.5 m/s² + 10 m/s²
→ 10.5 m/s²
Finally, substitute these values in the formula. We get,
workdone = 30 kg × 10.5 m/s² × 10 m
→ workdone = 30 × 105
∴ workdone = 3150 J
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