A cricket ball of mass 0.1 kg moving with a speed of 40 m s-1 is brought to rest by a player in 0.04 s. Find the average force applied by the player.
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Answers
Explanation:
mass of the ball (m) =100g or 0.1kg
initial velocity (u) =25
s
m
time (t) = 0.025sec.
final velocity (v) = 0
Initial momentum = mu
= 0.1×25
= 2.5
s
kgm
Final momentum = mv (which is equal to 0, since v is zero)
Change in momentum = mv-mu
= 0 - 2.5 = -2.5
s
kgm
Average force =
time
Changeinmomentum
= -
0.025
(2.5)
= -100
s
kgm
= -100 N
The negative sign actually shows that the force was applied
opposite to the direction of the motion of the ball.
So, the average force applied by the player will be 100
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Explanation:
HORT / DIRECT METHOD :-
F = \frac{mdv} {dt}F=
dt
mdv
F = 100^{-3} X \frac{40^{10}}{4 X 10^{-2}}F=100
−3
X
4X10
−2
40
10
F = 100 X \frac{10^{-2}}{ 10^{-2}} = 100 NF=100X
10
−2
10
−2
=100N
LONG METHOD :-
Concept used: 1) v = u + at (Ist Equation of Motion)
v - u = at
Here, v = final velocity
u = initial velocity
t = time
a = acceleration
2) F = ma (Newton's Second Law of Motion)
Here, m = mass of body
F = Force applied
Mass of ball = 100 g = 0.1 Kg
Initial speed (u) = 40 m/s
Final speed (v) = 0 m/s
Time (t) = 0.04 s
Acceleration (a) =
a =
a = - 1000 m/s²
F = ma
F = 0.1 × - 1000
F = - 100 N
∴ Force applied by the player = 100 N in the direction opposite to the ball thrown.