How Is Work Calculated When Lifting an Object with a Tension Force?

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To calculate the work done by the tension force when lifting a 100 kg mass upward with an acceleration of 1 m/s² over a distance of 5 m, the total tension must account for both gravitational force and the applied force. The gravitational force is 981 N (using 9.81 m/s²), and the applied force due to acceleration is 100 N, leading to a total tension of 1081 N. The work done by the tension force is calculated as 1081 N multiplied by 5 m, resulting in 5405 J. This calculation confirms that the correct approach involves adding the gravitational force and the force due to acceleration to find the total tension. Thus, the work done by the tension force is 5405 J.
NoHeart
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here's the question:
a mass of 100kg is pulled directly upward a distance of 5m by a rope with acceleration 1m/s^2. what is the work done by the tension force?

so the normal force is mg, 100kg*10m/s^2(gravitational constant given)= 1000N
the applied force is 100kg*1m/s^2= 100N
so is the tension 1000+100? if so the work done is 1100N*5m= 5500J
it's being pulled straight up, so the angle is 0, cos0=1

is it correct to add the two forces to find the total tension?
 
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Last edited:
NoHeart said:
here's the question:
a mass of 100kg is pulled directly upward a distance of 5m by a rope with acceleration 1m/s^2. what is the work done by the tension force?

so the normal force is mg, 100kg*10m/s^2(gravitational constant given)= 1000N
the applied force is 100kg*1m/s^2= 100N
so is the tension 1000+100? if so the work done is 1100N*5m= 5500J
it's being pulled straight up, so the angle is 0, cos0=1

is it correct to add the two forces to find the total tension?
SOLUTION HINTS:
From problem statement:
{Mass of Object} = m = (100 kg)
{Gravitational Force Downward on Object} = m*g = (100 kg)*(9.81 m/sec^2) = (981 N)
{Acceleration of Object Upwards} = a = (1 m/sec^2)
{Distance Object Pulled Upwards} = H = (5 m)
{Rope Tension Force Upward} = T

{NET Total Force on Object Upward} = m*a =
= {Rope Tension Force Upwards} - {Gravitational Force Downward} =
= T - m*g
::: ⇒ m*a = T - m*g
::: ⇒ T = m*a + m*g

{Work Done By Rope Tension} =
= {Rope Tension Force Upward}*{Distance Object Pulled Upwards} =
= T*H =
= {m*a + m*g}*H

::: ⇒ {Work Done By Rope Tension} = {m*a + m*g}*H
Evaluate above equation to determine work performed by Rope Tension.


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The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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