A block of mass 2 kg is initially at rest. A force is applied, and the block achieves a velocity of 3 m/s after traveling a distance of 1.5 m. What is the work done on the block?
Understand the Problem
We need to calculate the work done on a block. We know its mass, initial velocity (at rest), final velocity, and the distance it travels. We can use the work-energy theorem to find the work done.
Answer
$W = 90 \text{ J}$
Answer for screen readers
$W = 90 \text{ J}$
Steps to Solve
- Identify the given quantities
Mass, $m = 5.0 \text{ kg}$ Initial velocity, $v_i = 0 \text{ m/s}$ Final velocity, $v_f = 6.0 \text{ m/s}$ Distance, $d = 2.0 \text{ m}$
- Apply the work-energy theorem
The work-energy theorem states that the net work done on an object is equal to the change in its kinetic energy: $W = \Delta KE = KE_f - KE_i$
- Calculate the initial kinetic energy
Since the block starts at rest, the initial kinetic energy is zero: $KE_i = \frac{1}{2} m v_i^2 = \frac{1}{2} (5.0 \text{ kg}) (0 \text{ m/s})^2 = 0 \text{ J}$
- Calculate the final kinetic energy
$KE_f = \frac{1}{2} m v_f^2 = \frac{1}{2} (5.0 \text{ kg}) (6.0 \text{ m/s})^2 = \frac{1}{2} (5.0 \text{ kg}) (36 \text{ m}^2/\text{s}^2) = 90 \text{ J}$
- Calculate the work done
$W = KE_f - KE_i = 90 \text{ J} - 0 \text{ J} = 90 \text{ J}$
$W = 90 \text{ J}$
More Information
The work done on the block is equal to the change in kinetic energy, which is 90 Joules. The distance traveled is not needed to solve this problem using the work-energy theorem.
Tips
A common mistake is to use the distance given and kinematic equations to find the force and then calculate work as $W = Fd$. While this approach would also yield the correct answer, it is more complex and introduces more opportunities for error. Using the work-energy theorem is a more direct route. Another common mistake is to forget that the initial kinetic energy is zero when the object starts from rest.
Another mistake is using the wrong units or forgetting to convert units, though that's not applicable here.
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