a 0.140-kg baseball traveling 35.0 m/s strikes the catchers mitt, which, in bringing the ball to rest, recoils backward 11.0 cm. What was the average force applied by the ball on the glove?
Answers
Answer:
779.54N
Explanation:
Given,
m=0.14kg, v=35m/s, s=11cm = 0.11m
let, acceleration = a
Now,
we know that, v² = 2as
(35)² = 2a×0.11
1225 = 0.22×a
a = 5568.18m/s²
so,
F = ma
= 0.14kg × 5568.18m/s²
= 779.54N
Given : 0.140-kg baseball traveling 35.0 m/s strikes the catchers mitt, which, in bringing the ball to rest, recoils backward 11.0 cm.
To Find : The average force applied by the ball on the glove
Solution:
Work Done = Change in Kinetic energy
Word done = Fd
d = 11 cm = 0.11 m
F = ?
m = 0.140 kg
Change in Kinetic energy = (1/2)m(0)² - (1/2)m(35)²
= -(1/2)(0.14)(35)²
= -85.75 Joule
85.75 = F(0.11)
=> F = 779.545 N
average force applied by the ball on the glove = 779.545 N
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