Answer:
a) 24 J
b) Gravitational Force
c) 45 J
d) 0
e) 6.782m/s
Explanation:
a) m = 3kg
v = 4m/s
h = 1.5m
KE = ?
0.5 * 3 * 16 = 24J
b) Gravitational force
c) F = ma = 3 * 10 = 30N
Work done = Force * distance = 30 * 1.5 = 45J
d) Final Kinetic Energy of the ball is zero because the ball eventually stops moving
e) velocity of ball as it strikes the ground = v
[tex]v^{2} - u^{2} = 2as[/tex] where
v is the velocity as it strikes the ground
u is the initial velocity
a is acceleration
s is the distance
Now since the ball is thrown downwards, a is positive because the velocity of the ball is increasing as the gravitational force acts on it
u = 4m/s
a = 10
s = 1.5
=> [tex]v = \sqrt{2as + u^{2} }[/tex]
= [tex]\sqrt{(2*10*1.5) + 16}[/tex]
= [tex]\sqrt{46} = 6.782m/s[/tex]
(a) The kinetic energy of the ball as it leaves the thrower’s hand is 24 J.
(b) The force doing work on the ball as it falls is gravitational force.
(c) The work done on the ball as it falls is 44.1 J.
(d) The final kinetic energy of the ball is zero when it hits the ground.
(e) The velocity of the ball as it strikes the ground is 6.74 m/s.
The given parameters:
The kinetic energy of the ball as it leaves the thrower’s hand is calculated as follows;
[tex]K.E = \frac{1}{2} mv^2\\\\K.E = \frac{1}{2} \times 3 \times 4^2\\\\K.E = 24 \ J[/tex]
The force doing work on the ball as it falls is gravitational force
The work done on the ball as it falls is calculated as follows;
W = mgh
W = 3 x 9.8 x 1.5
W = 44.1 J
The final kinetic energy of the ball is zero when it hits the ground.
The velocity of the ball as it strikes the ground is calculated as follows;
[tex]v^2 = u^2 + 2gh\\\\v = \sqrt{4^2 + 2(9.8)(1.5)} \\\\v = 6.74 \ m/s[/tex]
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