Question

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

Answer 1

F=ma
=300×0.5
=150N
S=10m
W=fs
=150×10
1500j

Answer 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