Ms. Fizics pushed a large crate of textbooks (75 kg) down the hall to her office. She pushed as hard as she could, and the crate accelerated down the hall at a rate of
. There was a coefficient of kinetic friction of 0.48 between the crate and the floor.
a) Draw a free-body diagram for the crate. [C /2]
b) Determine the net force on the crate. [A /1]
c) Determine the crate's normal force. [A /1]
d) Determine the force of friction on the crate. [A /1]
e) Determine the force exerted by Ms. Fizics. [A /1]
f) How much work has she done when she travels 12 m in the same direction as the applied force? [A /2]
g) Determine the final velocity of the crate after 12 m, assuming the crate starts from rest. [A /2]
a) The free-body diagram for the crate would show the following forces:
- A downward force of gravity (mg), where m is the mass of the crate (75 kg) and g is the acceleration due to gravity (9.8 m/s^2)
- An upward normal force (N) from the floor, perpendicular to the surface
- A force of kinetic friction (f) acting in the opposite direction of motion
b) The net force on the crate can be determined by subtracting the force of friction from the applied force. Since the crate is accelerating, the applied force must be greater than the force of friction.
c) The crate's normal force is equal to the force of gravity (mg) since the crate is on a flat surface and not accelerating vertically.
d) The force of friction can be calculated by multiplying the coefficient of kinetic friction (0.48) by the normal force.
e) The force exerted by Ms. Fizics is equal in magnitude and opposite in direction to the force of kinetic friction.
f) The work done by Ms. Fizics can be calculated by multiplying the force she exerts (equal to the force of kinetic friction) by the distance she travels in the same direction (12 m).
g) The final velocity of the crate after traveling 12 m can be calculated using the equation vf^2 = vi^2 + 2ad, where vi is the initial velocity (0 m/s since the crate starts from rest), a is the acceleration (calculated from the net force divided by the crate's mass), and d is the distance traveled (12 m).
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A ball is dropped from a balcony.
a) Give the energy transformation equation for one complete ball bounce. Drawing labelled diagrams can help but is not mandatory. [T /3]
b) Explain how you could determine the efficiency with which the mechanical energy of a ball bounce is conserved. [T /2]
a) The energy transformation equation for one complete ball bounce can be described as follows:
Potential energy (at the highest point) -> Kinetic energy (during free fall) -> Elastic potential energy (during compression of the ball) -> Kinetic energy (during rebound) -> Potential energy (at the highest point for the subsequent bounce)
This sequence represents the transformation of energy between potential energy and kinetic energy during the different stages of the ball's motion.
b) To determine the efficiency with which the mechanical energy of a ball bounce is conserved, one can compare the initial potential energy of the ball (at the highest point) with the mechanical energy after the bounce. The mechanical energy after the bounce is the sum of the ball's kinetic energy and potential energy at the highest point.
Efficiency (%) = (Mechanical energy after bounce / Initial potential energy) x 100
If the mechanical energy after the bounce is equal to the initial potential energy, then the efficiency would be 100%, indicating that all the initial potential energy was conserved during the bounce. However, in reality, some energy is lost to other forms, such as heat or sound, as the ball compresses and rebounds. Therefore, the efficiency will be less than 100%, indicating that some energy was not conserved during the bounce.
a) To draw a free-body diagram for the crate, we need to consider all the forces acting on it. These forces include:
- The weight of the crate acting downward (mg, where m is the mass of the crate and g is the acceleration due to gravity).
- The normal force acting upward from the floor.
- The force of friction acting in the opposite direction of motion.
b) The net force on the crate is equal to the applied force minus the force of friction. We can calculate it using the equation:
Net force = Applied force - Force of friction
c) The normal force is the force exerted by the floor on the crate to balance the weight of the crate. It is always perpendicular to the surface and equal in magnitude and opposite in direction to the weight of the crate.
d) The force of friction can be calculated using the equation:
Force of friction = coefficient of friction * normal force
e) The force exerted by Ms. Fizics is the applied force that she pushed with.
f) The work done is calculated using the equation:
Work = force * distance * cos(angle)
In this case, the force is the applied force, the distance is 12 m, and the angle is 0 degrees (since the force and displacement are in the same direction).
g) To determine the final velocity of the crate after 12 m, assuming it starts from rest, we can use the equation of motion:
vf^2 = vi^2 + 2ad
where vf is the final velocity, vi is the initial velocity (0 in this case), a is the acceleration, and d is the distance traveled. Rearranging the equation, we can solve for vf.