Question

21. A body of mass 2 kg is at a height of 10 m above the ground. By applying work-energy principle : (a) Calculate speed of the ball at a height of 6 m from the ground, when it falls freely from top. (b) Calculate speed of the ball, when it reaches the ground after free fall from a height of 10 m. (c) How much work is done by gravitational force in bringing the ball at a height of 10 m from the ground ? (d) State whether the work done in (c) will be positive or negative. (Take g=10 ms)​

Answer 1

Answer:

Given : m=0.2kg,g=10ms−2,h1=10m,h2=6m.

Loss in P.E. =mg(h1−h2)

=0.2×10×10(10−6)

=0.2×10×4=8J

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Hope this helps:)

Answer 2

Explanation:

- two formula is use in this

work = mgh ( in first three )

K.E = 1/2 mv^2 ( in first two )

ANSWER:

1) given

m = 2kg

g = 10 ms^-1

h1 = 10 m and h2 = 6 m

so ,

work = mg(h1-h2)

work = 2×10×(10-6)

work = 2×10×4

work = 80 J

K.E = 1/2 mv^2 = 80,

K.E = 80× 2 v^2 = 80,

K.E = ✓80 = V

V = 8.9ms^-1

SECOND PART WILL BE SAME IF YOU TRY YOU CAN LEARN . TIP - HEIGHT WILL REMAIN SAME IN THIS READ QUESTION CARE FULLY .

3) WORK = mgh = 2×10×10= 200J

4) when the object lifted up , the work done by the gravitational force is negative.

I HOPE ITS HELPFUL FOR YOU

THANKU