Does work done on an object depend on the force applied? EX:A 50kg barbell is lifted 1.5 meters. How much work is done?

I am confused because I do not know how much force was used to lift the barbell (And I don't know the acceleration). I know the gravitational force but how do I calculate the net force?

My friend believes that we do not have enough information

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    $\begingroup$ For the problem you provided, don't you just multiply the gravitational force by 1.5 meters to figure out the work? Work done against gravity is always just mgh. $\endgroup$ – Kane Billiot Nov 2 '17 at 4:17
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    $\begingroup$ Check out this question. I think you will find your answer there. $\endgroup$ – Kane Billiot Nov 2 '17 at 4:18

One of the Theorems relating work and energy is $$W_{C,A \to B} = - \Delta U,$$ where $W_{C, A \to B}$ represents the work done by the conservative forces between two points $A$ and $B$ and $\Delta U$ represents the change in the potential energy.

The work done by a force between two points $A$ and $B$ is defined as $$W_{A \to B} = \int_A^B \vec{F} \cdot \text{d}\vec{s}.$$

Since every force related to a potential (by the formula $\vec{F} = - \nabla U$) is a conservative force, it comes easily that:

$$\vec{F} = - \nabla U$$ $$\int_A^B \vec{F} \cdot \text{d}\vec{s} = - \int_A^B \nabla U \cdot \text{d}\vec{s}$$ $$ W_{C, A \to B} = - \left[ U(B) - U(A) \right]$$ $$ \therefore W_{C,A \to B} = - \Delta U$$

Since, when close to the ground, $U(h) = mgh$, you just have to calculate $U(A) - U(B) = mg \cdot h(A) - mg \cdot h(B)$.

$$W = 50 \cdot g \cdot 0 - 50 \cdot g \cdot 1.5 = - 75 \cdot g$$ $$ \therefore W = -750 \text{ J}$$

As you can see, the work is negative, since the weight force points downwards and the displacement points upwards.


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