How much work must be performed by a hand or other such agent to move the 8kg block very slowly 10 cm to the right if the attached spring was initally relaxed? Let k = 200 N/m and mK = 0.5.
Show the equation and the plug the numbers into the equation and then the answer at the end
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To calculate the work done by a hand or agent to move the 8kg block slowly 10 cm to the right, we need to consider the work done against the force of the spring.
The equation for the work done by a force is given by:
Work = Force * Distance * cos(theta)
In this case, the force is the force exerted by the spring, the distance is the displacement of the block, and theta is the angle between the direction of the force and the displacement. Since the block is moved directly to the right, theta will be 0 degrees, and the cos(theta) term will equal 1.
The force exerted by the spring can be calculated using Hooke's Law, which states that the force exerted by a spring is proportional to the displacement and the spring constant:
Force = -k * displacement
where k is the spring constant. Since the block is moved to the right, the displacement will be positive.
For this problem, k is given as 200 N/m and the displacement is given as 10 cm (which needs to be converted to meters).
First, let's convert the displacement to meters:
10 cm = 10/100 = 0.1 m
Now, we can calculate the force exerted by the spring:
Force = -k * displacement = -200 * 0.1 = -20 N
Finally, we can calculate the work done:
Work = Force * Distance = -20 * 0.1 = -2 J (Joules)
Note that the negative sign indicates that the work is done against the force of the spring. In this case, the hand or agent needs to exert a force in the opposite direction of the spring force to move the block.
So, the work done by the hand or agent to move the 8kg block very slowly 10 cm to the right, with the given parameters, is -2 Joules.