Questions: A 11 kg object is moving in a straight-line with an initial speed of 8.8 m / s. What work (in J) must be done on the object for the object's speed to decrease to 3.3 m / s?

A 11 kg object is moving in a straight-line with an initial speed of 8.8 m / s. What work (in J) must be done on the object for the object's speed to decrease to 3.3 m / s?

Transcript text: A 11 kg object is moving in a straight-line with an initial speed of $8.8 \mathrm{~m} / \mathrm{s}$. What work (in J) must be done on the object for the object's speed to decrease to $3.3 \mathrm{~m} / \mathrm{s}$ ?

Solution

Solution Steps

Step 1: Identify the initial and final kinetic energies

The kinetic energy (KE) of an object is given by the formula: \[ KE = \frac{1}{2} m v^2 \] where $ m $ is the mass and $ v $ is the velocity.

First, we calculate the initial kinetic energy ($ KE_{\text{initial}} $): \[ KE_{\text{initial}} = \frac{1}{2} \times 11 \, \text{kg} \times (8.8 \, \text{m/s})^2 \]

Step 2: Calculate the initial kinetic energy

\[ KE_{\text{initial}} = \frac{1}{2} \times 11 \times 77.44 \] \[ KE_{\text{initial}} = 5.5 \times 77.44 \] \[ KE_{\text{initial}} = 425.92 \, \text{J} \]

Step 3: Calculate the final kinetic energy

Next, we calculate the final kinetic energy ($ KE_{\text{final}} $): \[ KE_{\text{final}} = \frac{1}{2} \times 11 \, \text{kg} \times (3.3 \, \text{m/s})^2 \] \[ KE_{\text{final}} = \frac{1}{2} \times 11 \times 10.89 \] \[ KE_{\text{final}} = 5.5 \times 10.89 \] \[ KE_{\text{final}} = 59.895 \, \text{J} \]

Step 4: Determine the work done

The work done on the object is the change in kinetic energy: \[ W = KE_{\text{final}} - KE_{\text{initial}} \] \[ W = 59.895 \, \text{J} - 425.92 \, \text{J} \] \[ W = -366.025 \, \text{J} \]