Question

1. a 3 kg object is moving in the +x-direction with a speed of 2 m/s at x=0. the net force acting on the object is force \\( \vec{f} \\), seen in the graph. what is its kinetic energy at x=7 m?

(a)12 j
(b)6 j
(c)0 j
(d)18 j
(e)14 j

1. a stone is dropped from the edge of a cliff. which of the following graphs best represents the stones kinetic energy ke as a function of time t?

Explanation:

To solve the problem of finding the kinetic energy of a 3 kg object at \( x = 7 \, \text{m} \), we use the work - energy theorem, which states that the work done by the net force on an object is equal to the change in its kinetic energy, \( W=\Delta KE=KE_f - KE_i \).

Step 1: Calculate the initial kinetic energy

The formula for kinetic energy is \( KE=\frac{1}{2}mv^{2} \). Given that the mass of the object \( m = 3\space kg \) and the initial velocity \( v_i=2\space m/s \) (at \( x = 0 \)):

$$ KE_i=\frac{1}{2}\times3\times(2)^{2}=\frac{1}{2}\times3\times 4 = 6\space J $$

Step 2: Calculate the work done by the force from \( x = 0 \) to \( x = 7 \, \text{m} \)

The work done by a variable force is equal to the area under the force - displacement (\( F - x \)) graph. The \( F - x \) graph is a trapezoid (or we can break it into a rectangle and a triangle).

• From \( x = 0 \) to \( x = 5 \, \text{m} \), the force \( F = 2\space N \). The area of this rectangular part is \( A_1=F\times\Delta x_1=2\times(5 - 0)=10\space J \)
• From \( x = 5 \, \text{m} \) to \( x = 7 \, \text{m} \), the force varies from \( F = 2\space N \) to \( F = 0\space N \). The area of this triangular part is \( A_2=\frac{1}{2}\times F_{avg}\times\Delta x_2 \). The average force \( F_{avg}=\frac{2 + 0}{2}=1\space N \) and \( \Delta x_2=7 - 5 = 2\space m \), so \( A_2=\frac{1}{2}\times2\times(7 - 5)=2\space J \)

The total work done \( W=A_1 + A_2=10+2 = 12\space J \)

Step 3: Calculate the final kinetic energy

Using the work - energy theorem \( W=KE_f-KE_i \), we can solve for \( KE_f \):

$$ KE_f=KE_i + W $$

Substitute \( KE_i = 6\space J \) and \( W = 12\space J \) into the formula:

$$ KE_f=6 + 12=18\space J $$

Answer:

\( 18\space J \) (corresponding to option d)