Question

2. A 3 kg ball is thrown downward at 4 m/s from a height of 1.5 m. a. What is the kinetic energy of the ball as it leaves the thrower’s hand? b. What force is doing work on the ball as it falls? c. How much work is done on the ball as it falls? d. What is the final kinetic energy of the ball? e. What is the velocity of the ball as it strikes the ground?

Answer 1

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

$v^{2} - u^{2} = 2as$ 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

=>  $v = \sqrt{2as + u^{2} }$

= $\sqrt{(2*10*1.5) + 16}$

= $\sqrt{46} = 6.782m/s$