Questions: There is a 140 Kg car that is accelerating to the right at a rate of 2.4 m / s^2, in front of the car is a 3.4 kg mass that is "stuck" to the car's front because of the system's motion. As the car is moving there is also a wind blowing to the left exerting a pressure of 290 Pa. If wind does not affect the car's motion and the mass is cubed in shape with sides that are 0.20 m in length, what is the normal force between the car and the mass in Newtons and rounded to two decimal places?

Solution Steps

The mass experiences the following forces:

• Gravity (Fg): Acting downwards with a magnitude of \(mg\), where \(m\) is the mass (3.4 kg) and \(g\) is the acceleration due to gravity (9.8 m/s²).
• Normal force (Fn): Acting upwards from the car's surface. This is the force we need to determine.
• Wind force (Fw): Acting to the left due to the wind pressure. This force is given by \(P \cdot A\), where \(P\) is the pressure (290 Pa) and \(A\) is the area of the mass facing the wind. Since the mass is a cube with sides of 0.20 m, the area is \(0.20^2 = 0.04 \mathrm{~m}^2\).
• Force due to acceleration (Fa): Acting to the right due to the car's acceleration. This force has a magnitude of \(ma\), where \(a\) is the car's acceleration (2.4 m/s²).

Since the mass is not moving vertically, the net force in the vertical direction is zero. This means the normal force must balance the gravitational force:

\(F_n = F_g = mg = (3.4 \mathrm{~kg})(9.8 \mathrm{~m/s}^2) = 33.32 \mathrm{~N}\)

The problem states that the wind doesn't affect the motion, and the mass is "stuck" to the car. This implies there must be a frictional force between the mass and the car preventing the mass from sliding off. However, since we are only asked about the normal force, we don't need to consider horizontal forces for this problem.

Final Answer