Question

questions 9 and 10 refer to the diagram which depicts two pucks on a frictionless table. puck ii is four times as massive as puck i. starting from rest, the pucks are pushed across the table by two equal forces.

1. which puck will have the greatest amount of work done on upon reaching the finish line?

a. i
b. ii
c. they both have the same amount.
d. cannot answer with the information provided.

1. which puck will have the greatest kinetic energy upon reaching the finish line?

a. i
b. ii
c. they both have the same amount.
d. cannot answer with the information provided.

Explanation:

Question 9

Step1: Recall the work formula

Work is calculated as \( W = F \cdot d \), where \( F \) is the force and \( d \) is the displacement in the direction of the force.

Step2: Analyze the forces and displacements

Both pucks are pushed with equal forces (\( F \)). We need to find the displacement of each puck. Using Newton's second law \( F = ma \), for Puck I (mass \( m \)): \( a_1=\frac{F}{m} \), for Puck II (mass \( 4m \)): \( a_2 = \frac{F}{4m} \). Using the kinematic equation \( d=v_0t+\frac{1}{2}at^2 \), with \( v_0 = 0 \), so \( d=\frac{1}{2}at^2 \). Let the time taken for Puck I to reach the finish line be \( t_1 \) and for Puck II be \( t_2 \). The distance to the finish line is the same, let's call it \( D \). So \( D=\frac{1}{2}a_1t_1^2=\frac{1}{2}a_2t_2^2 \). Substituting \( a_1 \) and \( a_2 \): \( \frac{1}{2}\cdot\frac{F}{m}\cdot t_1^2=\frac{1}{2}\cdot\frac{F}{4m}\cdot t_2^2 \), simplifying gives \( t_2^2 = 4t_1^2 \), so \( t_2 = 2t_1 \). Now, calculate work for Puck I: \( W_1=F\cdot d_1 \), but \( d_1=\frac{1}{2}a_1t_1^2=\frac{1}{2}\cdot\frac{F}{m}\cdot t_1^2 \). For Puck II: \( d_2=\frac{1}{2}a_2t_2^2=\frac{1}{2}\cdot\frac{F}{4m}\cdot(2t_1)^2=\frac{1}{2}\cdot\frac{F}{4m}\cdot4t_1^2=\frac{1}{2}\cdot\frac{F}{m}\cdot t_1^2 \). So \( d_1 = d_2 \) (since both travel distance \( D \)). Since \( F \) is equal and \( d \) is equal, from \( W = Fd \), \( W_1 = W_2 \).

Step1: Recall kinetic energy formula

Kinetic energy is \( KE=\frac{1}{2}mv^2 \). We can also use the work - energy theorem, which states that the work done on an object is equal to its change in kinetic energy (\( W=\Delta KE \)). From question 9, we know that the work done on both pucks is the same ( \( W_1 = W_2 \) ). Since both start from rest, the initial kinetic energy is 0, so the final kinetic energy \( KE = W \).

Step2: Conclude the kinetic energy

Since the work done on both pucks is the same, their change in kinetic energy (and thus final kinetic energy, as initial \( KE = 0 \)) is the same.

Answer:

c. They both have the same amount.

Question 10