The force of attraction between a ball is F=.........×10^-¹¹
1 answer:
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The coefficient of friction is missing and it has a value of μ = 0.4
Answer:
a = 3.924 m/s²
Explanation:
I've attached the kinematic free body diagram.
Taking the sum of all upward and downward forces,
EFy = 0;
N1 + N2 - m_p•g - m_v•g = 0
N1 + N2 = m_p•g + m_v•g
Where;
N1 and N2 are the normal reactions at the wheels
m_p is the mass of the driver
m_v is the mass of the vehicle
g is the acceleration due to gravity.
Plugging in the relevant values in the question,we obtain;
N1 + N2 = (385 + 75) x 9.81
N1 + N2 = 4512.6N - - - (eq1)
Now, taking sum of all horizontal forces;
EFx = (m_p + m_v) x a
So,
μ(N1 + N2) = (mp + mv) x a
Thus,
0.4(N1 + N2) = (385 + 75)a
0.4(N1 + N2) = 460a
N1 + N2 = 1150a
From eq(1),N1 + N2 = 4512.6N
Thus,
1150a = 4512.6N
a = 4512.6/1150
a = 3.924 m/s²
Therefore, the acceleration, a = 3.924 m/s²
