QuestionJune 14, 2025

A 60.0 kg object is moving east at 8.00m/s and then slows down to 4.00m/s How much work was done? -1,440 J -480 1,440J 2,880J

A 60.0 kg object is moving east at 8.00m/s and then slows down to 4.00m/s How much work was done? -1,440 J -480 1,440J 2,880J
A 60.0 kg object is moving east at
8.00m/s and then slows down to 4.00m/s How much work was done?
-1,440 J
-480
1,440J
2,880J

Solution
4.7(192 votes)

Answer

-1,440 J Explanation 1. Identify the initial and final velocities The object initially moves at v_i = 8.00 \, \text{m/s} and slows down to v_f = 4.00 \, \text{m/s}. 2. Determine the mass of the object The mass of the object is given as m = 60.0 \, \text{kg}. 3. Calculate the initial kinetic energy The initial kinetic energy is calculated using the formula KE_i = \frac{1}{2} m v_i^2. Substituting the values, we get KE_i = \frac{1}{2} \times 60.0 \, \text{kg} \times (8.00 \, \text{m/s})^2 = 1920 \, \text{J}. 4. Calculate the final kinetic energy The final kinetic energy is calculated using the formula KE_f = \frac{1}{2} m v_f^2. Substituting the values, we get KE_f = \frac{1}{2} \times 60.0 \, \text{kg} \times (4.00 \, \text{m/s})^2 = 480 \, \text{J}. 5. Calculate the work done The work done on the object is the change in kinetic energy, which is W = KE_f - KE_i = 480 \, \text{J} - 1920 \, \text{J} = -1440 \, \text{J}.

Explanation

1. Identify the initial and final velocities<br /> The object initially moves at $v_i = 8.00 \, \text{m/s}$ and slows down to $v_f = 4.00 \, \text{m/s}$.<br />2. Determine the mass of the object<br /> The mass of the object is given as $m = 60.0 \, \text{kg}$.<br />3. Calculate the initial kinetic energy<br /> The initial kinetic energy is calculated using the formula $KE_i = \frac{1}{2} m v_i^2$. Substituting the values, we get $KE_i = \frac{1}{2} \times 60.0 \, \text{kg} \times (8.00 \, \text{m/s})^2 = 1920 \, \text{J}$.<br />4. Calculate the final kinetic energy<br /> The final kinetic energy is calculated using the formula $KE_f = \frac{1}{2} m v_f^2$. Substituting the values, we get $KE_f = \frac{1}{2} \times 60.0 \, \text{kg} \times (4.00 \, \text{m/s})^2 = 480 \, \text{J}$.<br />5. Calculate the work done<br /> The work done on the object is the change in kinetic energy, which is $W = KE_f - KE_i = 480 \, \text{J} - 1920 \, \text{J} = -1440 \, \text{J}$.
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