a) The work done to stop the ball is -133.1 J
b) The average force on the ball is [tex]-4.44\cdot 10^3 N[/tex]
Explanation:
a)
According to the work-energy theorem, the work done on an object is equal to the change in kinetic energy of the object, therefore:
[tex]W=\Delta K = K_f - K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2[/tex]
where
W is the work done
m is the mass of the object
u is its initial velocity
v is its final velocity
For the ball in this problem:
m = 0.169 kg
u = 39.7 m/s
v = 0
Therefore, the work done on the ball is
[tex]W=0-\frac{1}{2}(0.169)(39.7)^2=-133.1 J[/tex]
And the work is negative since the force applied to stop the ball is opposite to the direction of motion of the ball.
b)
The motion of the ball is a uniformly accelerated motion, so we can calculate the acceleration by using the following suvat equation:
[tex]v^2-u^2 = 2as[/tex]
where
v is the final velocity
u is the initial velocity
a is the acceleration
s is the distance covered by the ball
Here we have
v = 0
u = 39.7 m/s
s = 3.0 cm = 0.03 m
So, the acceleration is
[tex]a=\frac{v^2-u^2}{2s}=\frac{0-(39.7)^2}{2(0.03)}=-2.63\cdot 10^4 m/s^2[/tex]
And the negative sign is because it is a deceleration.
So, the force applied on the ball is:
[tex]F=ma=(0.169)(-2.63\cdot 10^4)=-4.44\cdot 10^3 N[/tex]
And the negative sign is because the direction of the force is opposite to the motion of the ball.
Learn more about work:
brainly.com/question/6763771
brainly.com/question/6443626
About forces:
brainly.com/question/11179347
brainly.com/question/6268248
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