Respuesta :
Answer:
a) k = 191.67 N\m
b) V = 22 m/s
c) t = 3.83s
d) 17.36m
e) 45.89 m
Explanation:
given:
F = 23 N
x = 12 cm = 0.12 m
mass of projectile, m = 5.7g = 5.7×10⁻³ kg
a) For a spring
F = kx
where,
F = Applied force
k = spring constant
x = change in in spring length
thus, we have
23 = k×0.12
or
k = 23/0.12
⇒ k = 191.67 N\m
b) From the conservation of energy between the start and the point of interest we have
[tex]\frac{1}{2}\times kx^{2}=\frac{1}{2}mV^{2}[/tex]
where,
V = velocity of the projectile at 1.37 m above the floor
[tex]\frac{1}{2}\times 191.67\times 0.12^{2}=\frac{1}{2}5.7\times 10^{-3}V^{2}[/tex]
V = 22 m/s
c) time in air (t)
applying the Newtons's equaton of motion
[tex]h = Vsin\Theta\times t +\frac{1}{2}a_{y}t^{2}[/tex]
substituting the values in the above equation we get
[tex]-1.37 = 22sin57^{\circ} \times t +\frac{1}{2}(-9.8)t^{2}[/tex]
or
[tex]4.9t^{2}-18.45t-1.37 = 0[/tex]
solving the qudratic equation for 't', we get
t = 3.83s
d) For maximum height ([tex]H_{max}[/tex])
we have from the equations of projectile motion
[tex]H_{max} =\frac{V^{2}sin^{2}\theta }{2g}[/tex]
substituting the values in the above equation we get
[tex]H_{max} =\frac{22^{2}sin^{2}57^{\circ} }{2\times 9.8}[/tex]
or
[tex]H_{max} =17.36 m[/tex]
the height with respect to the ground surface will be = 17.36m + 1.37 m 18.73m
e)For horizontal distance traveled (R) we have the formula
[tex]R = Vcos\theta\times t[/tex]
substituting the values in the above equation we get
[tex]R =22\times cos57^{\circ} \times 3.83[/tex]
or
[tex]R =45.89 m[/tex]
