Respuesta :
Explanation:
The given data is as follows.
F = 3.2 N, m = 18.2 kg,
t = 0.82 sec
(a) Formula for impulse is as follows.
I = Ft = [tex]\Delta P[/tex]
Ft = [tex]m(v_{f} - v_{i})[/tex]
or, [tex]v_{f} = \frac{Ft}{m} + v_{i}[/tex]
Putting the given values into the above formula as follows.
[tex]v_{f} = \frac{Ft}{m} + v_{i}[/tex]
= [tex]\frac{3.2 \times 0.82}{18.2} + 0[/tex]
= 0.144 m/s
Therefore, final velocity of the mass if it is initially at rest is 0.144 m/s.
(b) When velocity is 1.85 m/s to the left then, final velocity of the mass will be calculated as follows.
Ft = [tex]m(v_{f} - v_{i})[/tex]
or, [tex]v_{f} = \frac{Ft}{m} + v_{i}[/tex]
= [tex]\frac{3.2 \times 0.82 sec}{18.2} - 1.85[/tex]
= -1.705 m/s
Hence, we can conclude that the final velocity of the mass if it is initially moving along the x-axis with a velocity of 1.85 m/s to the left is 1.705 m/s towards the left.
Answer:
Explanation:
mass, m = 18.2 kg
time, t = 0.82 s
Force, F = 3.2 N right
initial velocity, u = 0 m/s
Let v is the final velocity.
(a) By the Newton's second law
F = ma
where a is the acceleration
3.2 = 18.2 x a
a = 0.176 m/s²
By using first equation of motion
v = u + at
v = 0 + 0.176 x 0.82
v = 0.144 m/s
(b) initial velocity, u = - 1.85 m/s
v = u + at
v = - 1.85 + 0.176 x 0.82
v = - 1.7 m/s
v = 1.7 m/s left
