Answer:
A) v = 6.93 m/s
B) v = 4.9 m/s
C) x_m = 0.015m
D) v_max = 5.2 m/s
Explanation:
We are given;
x = 6 cm = 0.06 m
k = 400 N
m = 0.03 kg
F = 6N
A) from work energy law, work dome by the spring on ball which now became a kinetic energy is;
Ws = K.E = ½kx²
Similarly, kinetic energy of ball is;
K.E = ½mv²
So, equating both equations, we have;
½kx² = ½mv²
Making v the subject gives;
v = √(kx²/m)
Plugging in the relevant values to give;
v = √((400 × 0.06²)/0.03)
v = √48
v = 6.93 m/s
B) If there is friction, the total work is;
Ws = ½kx² - - - (1)
Work of the ball is;
Wb = KE + Wf
So, Wb = ½mv² + fx - - - (2)
Combining both equations, we have;
½mv² + fx = ½kx²
Plugging in the relevant values, we have;
(½ × 0.03 × v²) + (6 × 0.06) = ½ × 400 × 0.06²
0.015v² + 0.36 = 0.72
0.015v² = 0.72 - 0.36
v² = 0.36/0.015
v = √24
v = 4.9 m/s
C) The speed is greatest where the acceleration stops i.e. where the net force on the ball is zero. (ie spring force matches 6.0N friction)
So, from F = Kx;
(x is measured into barrel from end where F = 0)
Thus; 6.0 = 400x
x_m = 6/400
x_m = 0.015m from the end after traveling 0.045m
D) Initial force on ball = (Kx - F) =
[(400 x 0.06) - 6.0] = 18N
Final force on ball = 0N
Mean Net force on ball = ½(18 + 0)
Mean met force, F_m = 9N
Net Work Done on ball = KE = 9N x 0.045m = 0.405 J
Thus;
½m(v_max)² = 0.405J
(v_max)² = 2 x 0.405/0.03
(v_max)² = 27
v(max) = √27
v_max = 5.2 m/s
