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JEE Main Physics Class 11 Work, Energy and Power Part 1 Questions
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© examsnet.com
Question : 9
Total: 100
A cricket ball of mass
0.15
k
g
is thrown vertically up by a bowling machine so that it rises to a maximum height of 20
m
after leaving the machine. If the part pushing the ball applies a constant force
F
on the ball and moves horizontally a distance of
0.2
m
while launching the ball, the value of
F
(
in
N
)
is
(
g
=
10
m
s
−
2
)
________.
[NA 3 Sep. 2020 (I)]
Your Answer:
Validate
Solution:
From work energy theorem,
W
=
F
.
s
=
Δ
K
E
=
1
2
m
v
2
Here
V
2
=
2
g
h
∴
F
.
s
=
F
×
2
10
=
1
2
×
15
100
×
2
×
10
×
20
∴
F
=
150
N
© examsnet.com
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