Work and energy for a lowered block

AI Thread Summary
The discussion focuses on calculating the work done by the cord's force and the weight of a block being lowered with a constant acceleration of g/5. Initial attempts to find the work done resulted in incorrect values due to miscalculations of the forces involved. A free body diagram was suggested to clarify the forces acting on the block, leading to a corrected tension force calculation. The correct tension force was determined to be 274.4 N, which will aid in accurately calculating the work done. Understanding the direction of forces and motion is crucial for solving the problem correctly.
Kpgabriel
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Homework Statement


A cord is used to vertically lower an initially stationary block of mass M = 35 kg at a constant downward acceleration of g/5. When the block has fallen a distance d = 6.5 m, find the work done by the cord's force on the block.
Find the work done by the weight of the block.
Find the kinetic energy of the block.
Find the speed of the block.

Homework Equations


W = F*d
F = m*a

The Attempt at a Solution


So I tried finding the first part of the question where I multiplied the mass, 35 kg, and multiplied it by the acceleration, 9.80/5= 1.96, to get F= 198N and then I multiplied the force by the distance to get the work and got 445.9 J but it is wrong.
 
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Note: Thread title changed to describe the question subject matter.

Have you drawn a the free body diagram for the situation? What forces are operating on the block? Is the force due to the cord the same as the net force on the block?
 
gneill said:
Note: Thread title changed to describe the question subject matter.

Have you drawn a the free body diagram for the situation? What forces are operating on the block? Is the force due to the cord the same as the net force on the block?
No, I do not think it is. So I drew out the free body diagram and I got that the Tension force of the cord, T - mg = ma. So T = ma + mg
For this I got 411.5 N
To solve for work I multiplied it by 6.5m and got 2675.4J which is wrong. I am not sure what I am doing.
 
Your equation describes a mass accelerating upwards with acceleration a. Your free body diagram should help you to sort out the directions of the vectors and the signs of the terms in your equation.
 
gneill said:
Your equation describes a mass accelerating upwards with acceleration a. Your free body diagram should help you to sort out the directions of the vectors and the signs of the terms in your equation.
So the Tension force would be equal to T= M*g - M*a and that is equal to 274.4N
 
That looks reasonable. Be sure to keep track of the directions of the forces acting versus the direction of the motion of the block.
 

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