I need someone to paraphrase this for me
1. Write a brief explanation of what happened to the coin before and after you removed the index card in Part 1 in terms of forces and Newton’s laws.
Answer:
When the coin is resting on the top of a beaker on an index card, it is not moving. There is a static friction between the coin and the index card. When I exerted force on the card and caused the card to move, the coin fell in to the cup. If this situation was to be explained using Newton’s laws,Newton’s first law of motion says that an object will remain at rest when the net force is zero. The fact that the coin stayed still on the card was because the net force was zero or balance. This is because the card is exerting the same magnitude but opposite direction force on the coin as the coin is exerting on it and so those two forces cancel out each other. The coin and the card exert force on each other because, for every reaction ( weight of the coin) there is an opposite but equal reaction (the force the card exerts to support the coin). But when force was exerted on the card, the net force was no longer zero and so the object moved. The coin fell to the glass and the card moved away in the direction the force was exerted. The coin accelerated downward because it’s weight was no longer supported by the card and it caused it to accelerate downward to the bottom of the glass.
(5 points)
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2. Based on your observations of the spring scale in Part 2, describe all the forces acting on an elevator as it moves from rest upward and stops at its floor.
Answer:
When the elevator moves up ward from rest, there is applied force by the strings that causes it to accelerate. But the applied force has to be greater than the force of gravity for it to accelerate upwards. If there are passengers in the elevator then their weight is applied towards the floor of the elevator and the elevator force also exerts a normal force on the passenger. When the elevator is accelerating upward the weight of the passengers appears to be greater than the real weight. When the elevator is coming to a rest at a floor, then it begins to decelerate and acceleration is less than the acceleration due to gravity. The weight of the passenger here appears to be less than the real weight.
(8 points)
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3. Use your data from Part 3 and Newton’s laws to explain why the force meter measures a force if the cart is moving at a constant velocity. You should cover the following points.
· State Newton’s first law.
· Identify the forces acting on the cart.
· Describe the relationships between applied force and mass and between applied force and weight as linear (from data or mathematical derivation by applying Newton’s first law).
· Explain why the data are linear in terms of Newton’s first law.
· Graph F versus mass + cart.
Answer:
Newton’s first law of motion says that an object at rest will remain at rest if the net force acting on it is zero and an object in motion will continue to move in a constant velocity if the net force acting on it is zero. The cart was moving at a constant velocity but still I got a value different from zero when I measured it with the force meter. This is because I was measuring the applied force only but if I calculated the net force by subtracting or adding other forces acting on the cart, I would probably get a zero net force. Forces acting on the cart were the applied force I was measuring, friction in opposite direction of the applied force, the weight of the cart on the table and the table’s force on the cart that was in opposite direction called normal force. This force is equal to the weight of the cart.
Mass (g) | Mass + Cart (g) | Mass + Cart (kg) | Weight (N) | Force (N) |
100 | 151 | 0.15 | 1.47 | 0.15 |
200 | 251 | 0.25 | 2.45 | 0.25 |
300 | 351 | 0.35 | 3.43 | 0.5 |
400 | 451 | 0.45 | 4.41 | 0.75 |
500 | 551 | 0.55 | 5.39 | 1 |
Mass of the cart = 51 grams
As more mass was added on the cart, the more applied force need. This means that applied force had a linear relationship with mass as both were increasing together. Also the weight and applied force has linear relationship with the weight of the cart because they were also both increasing together. They are linear because according to Newton’s second law of motion mass is proportional with force.
PICTURE
(8 points)
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4. For Part 4, describe the motion of the ball as it moves from the top of the ramp and comes to rest in the cup. Explain how the motion changes in terms of the velocity, distance, acceleration, and force that you measured. Use your data and Newton’s laws to calculate these values and explain the situation. Include the tables you used to calculate average initial velocity and average displacement. Also show your work for your calculations of acceleration and net force.
Here are the kinematic equations:
PICTURE 4
Answer:
when the ball was rolling down the ramp, the force of gravity was acting on it and therefore it was decelerating. Once the ball reached the end, it collided with the cup and moved a few distance and came to stop. The ball had a constant acceleration which means the velocity was not constant. The acceleration of the ball can be calculate using a kinematic formula v_{f}^{2} = v_{i}^{2} + 2aΔd. This formula can be modified to a = v_{f}^{2} –v_{i ^2 / 2Δd. }
Final velocity was zero and the first velocity was 1 m/s. Moreover the change in displacement was 0.36 meters
a =( 0m/s)^2 – (1m/s )^2 / 2* 0.36
a= -1.4 m/s^2
Time (s) | Distance (m) | Velocity (m/s) |
0.93 | 1 | 1.08 |
1 | 1 | 1 |
1 | 1 | 1 |
1 | 1 | 1 |
1.07 | 1 | 0.93 |
Average velocity (m/s) = 1. 0 m/s
∆d (cm) | ∆d (m) | |
Trial 1 | 34 | 0.34 |
Trial 2 | 36 | 0.36 |
Trail 3 | 36 | 0.36 |
Trial 4 | 38 | 0.38 |
Average ∆d(m) = 0.36 m
The net force of the collision of the ball with cup can also be calculated using the formula
Fnet= mass x acceleration
Fnet = ?
a = 1.4
m= 0.67kg
Fnet = 0.67kg * 1.4 m/s^2
Fnet = 0.94 N