1. The image shows a man pulling a truck. If it does not show up for you for whatever reason, this is a description of the image.
Assume it takes 2500 N for him to pull the truck along the road at a constant speed.
a. Find the net force acting on the truck. (1.5 points)
b. Find the force of friction acting on the truck. (1.5 points)
2. Find the acceleration of a 10 N on a 10kg mass. How does it compare to the acceleration if you double the force on an object that is double the mass? Support your answer with work. (3 points)
3. How much pressure are you exerting on the ground when you are standing on one foot? (Assume your shoes are rectangular shaped.)
* convert to S.I. Units. (6 points)
1a. Fnet = M*a.
Fnet = M*0,
Fnet = 0.
b. F - Fk = M*a.
2500 - Fk = M*0 = 0,
Fk = 2500 N. = Force of kinetic friction.
2. F = M*a.
a = F/M
The Eq, a = F/M, states that if we double M, the acceleration will be cut in
half.
To find the answers to these physics questions, we will use the formulas related to force, friction, acceleration, and pressure.
1. Net force acting on the truck:
The net force is the resultant force acting on an object. It is determined by subtracting the force of friction from the applied force. In this case, the applied force is 2500 N. If we assume there is no other force acting on the truck, then the net force will be equal to the applied force. Therefore, the net force acting on the truck is 2500 N.
2. Acceleration of a 10 N mass:
We can use Newton's second law of motion to find the acceleration. Newton's second law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. The formula is: Acceleration (a) = Net Force (F_net) / Mass (m). Substituting the given values, we get: a = 10 N / 10 kg = 1 m/s^2.
Acceleration when force is doubled on an object with double the mass:
If we double the force on an object and also double its mass, the acceleration of the object will remain the same. This is because when both the force and mass are doubled, the ratio between them remains the same, resulting in the same acceleration. This can be mathematically shown as follows:
Initial force (F1) = 10 N
Initial mass (m1) = 10 kg
Initial acceleration (a1) = F1 / m1 = 10 N / 10 kg = 1 m/s^2
Double force (F2) = 2 * F1 = 2 * 10 N = 20 N
Double mass (m2) = 2 * m1 = 2 * 10 kg = 20 kg
New acceleration (a2) = F2 / m2 = 20 N / 20 kg = 1 m/s^2
As you can see, despite doubling the force and the mass, the acceleration remains the same.
3. Pressure exerted on the ground when standing on one foot:
To calculate the pressure, we use the formula: Pressure (P) = Force (F) / Area (A). Assuming the shape of your shoe is rectangular, we need to determine the area of contact with the ground.
Let's say the length of your shoe (L) is 25 cm (converted to meters, 0.25 m), the width (W) is 10 cm (converted to meters, 0.1 m), and your weight (which is the force exerted) is 600 N.
Area of contact (A) = Length (L) * Width (W) = 0.25 m * 0.1 m = 0.025 m^2
Pressure (P) = Force (F) / Area (A) = 600 N / 0.025 m^2 = 24000 Pa (Pascal). This is the pressure exerted on the ground when standing on one foot.
Note: Pascal (Pa) is the SI unit for pressure.