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Answer :
1. The speed of the box after being pushed by a 12 N force for 8 m would be approximately 5.7 m/s.
2. The box would move for approximately 2.04 seconds before stopping.
3. The cart would move with a speed of approximately 62.5 m/s when it reaches the bottom of the hill.
1. To determine the speed of the 6 kg box after being pushed by a 12 N force for a distance of 8 m, we can use Newton's second law of motion. This law states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.
First, we need to find the acceleration of the box. We can use the formula: force = mass × acceleration. Rearranging the formula, we get acceleration = force ÷ mass.
So, the acceleration of the box would be: 12 N ÷ 6 kg = 2 m/s^2.
Next, we can use the equation: final velocity^2 = initial velocity^2 + 2 × acceleration × distance.
Since the box is at rest initially, the initial velocity is 0 m/s. Plugging in the values, we get:
final velocity^2 = 0^2 + 2 × 2 m/s^2 × 8 m = 32 m^2/s^2.
Taking the square root of both sides, we find that the final velocity of the box is approximately 5.7 m/s.
Therefore, the speed of the box after being pushed by a 12 N force for 8 m would be approximately 5.7 m/s.
2. To find how long the 8 kg box, moving with a speed of 6 m/s, would move before stopping on a bumpy surface with a coefficient of friction of 0.3, we need to consider the force of friction acting on the box.
The force of friction can be calculated using the equation: force of friction = coefficient of friction × normal force.
The normal force is equal to the weight of the box, which is given by the equation: weight = mass × acceleration due to gravity. Here, the acceleration due to gravity is approximately 9.8 m/s^2.
So, the weight of the box is: 8 kg × 9.8 m/s^2 = 78.4 N.
Now, we can calculate the force of friction: force of friction = 0.3 × 78.4 N = 23.52 N.
To determine the acceleration of the box, we can use the formula: force = mass × acceleration. Rearranging the formula, we get acceleration = force ÷ mass.
So, the acceleration of the box is: 23.52 N ÷ 8 kg = 2.94 m/s^2.
Next, we can use the equation: final velocity = initial velocity + (acceleration × time).
Since the box eventually comes to a stop, the final velocity is 0 m/s. Plugging in the values, we get:
0 = 6 m/s + (2.94 m/s^2 × time).
Rearranging the equation, we find: time = -6 m/s ÷ 2.94 m/s^2.
Therefore, the box would move for approximately 2.04 seconds before stopping.
3. To determine how fast the 1500 lb cart would move when it reaches the bottom of the hill with a height of 350 m, we can use the principle of conservation of energy.
The potential energy at the top of the hill is given by the equation: potential energy = mass × acceleration due to gravity × height.
Converting the mass of the cart from pounds to kilograms: 1500 lb × 0.4536 kg/lb = 680.4 kg.
Now, we can calculate the potential energy: potential energy = 680.4 kg × 9.8 m/s^2 × 350 m.
Next, we can equate the potential energy at the top of the hill to the kinetic energy at the bottom of the hill using the equation: potential energy = kinetic energy.
The kinetic energy is given by the equation: kinetic energy = 0.5 × mass × velocity^2.
Assuming the velocity at the bottom of the hill is v, we have:
680.4 kg × 9.8 m/s^2 × 350 m = 0.5 × 680.4 kg × v^2.
Simplifying the equation, we find:
v^2 = (2 × 680.4 kg × 9.8 m/s^2 × 350 m) ÷ 680.4 kg.
Taking the square root of both sides, we get:
v = √(2 × 9.8 m/s^2 × 350 m) ≈ 62.5 m/s.
Therefore, the cart would move with a speed of approximately 62.5 m/s when it reaches the bottom of the hill.
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