A 100 kg box is decelerated from 10m/s to 8m/s. What is the magnitude of work needed to slow the box? a 1800J b 100J c 3600J d 6001

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A 100 kg box is decelerated from 10m/s to 8m/s. What is the magnitude of work needed to slow the box? a 1800J b 100J c 3600J d 6001

Answer 1

To solve this problem, we will use the work-energy principle, which states that the work done on an object is equal to the change in its kinetic energy. The formula for kinetic energy (\(KE\)) is \(\frac{1}{2}mv^2\), where \(m\) is the mass and \(v\) is the velocity of the object.**Step 1:** Calculate the initial kinetic energy (\(KE_{initial}\)) using the initial velocity (\(10 \mathrm{~m/s}\)).\[ KE_{initial} = \frac{1}{2} \times 100 \mathrm{~kg} \times (10 \mathrm{~m/s})^2 \]\[ KE_{initial} = \frac{1}{2} \times 100 \times 100 \]\[ KE_{initial} = 50 \times 100 \]\[ KE_{initial} = 5000 \mathrm{~J} \]**Step 2:** Calculate the final kinetic energy (\(KE_{final}\)) using the final velocity (\(8 \mathrm{~m/s}\)).\[ KE_{final} = \frac{1}{2} \times 100 \mathrm{~kg} \times (8 \mathrm{~m/s})^2 \]\[ KE_{final} = \frac{1}{2} \times 100 \times 64 \]\[ KE_{final} = 50 \times 64 \]\[ KE_{final} = 3200 \mathrm{~J} \]**Step 3:** Determine the work done (\(W\)) by finding the change in kinetic energy.\[ W = KE_{final} - KE_{initial} \]\[ W = 3200 \mathrm{~J} - 5000 \mathrm{~J} \]\[ W = -1800 \mathrm{~J} \]The negative sign indicates that work is done to decelerate the box, but since the question asks for the magnitude of work, we consider the absolute value.**Answer**: The magnitude of work needed to slow the box is \( 1800 \mathrm{~J} \), which corresponds to option (a) \( 1800 \mathrm{~J} \).

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A 100 kg box is decelerated from 10m/s to 8m/s. What is the magnitude of work needed to slow the box? a 1800J b 100J c 3600J d 6001

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