0
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ Check out this question. I think you will find your answer there. $\endgroup$ – Kane Billiot Nov 2 '17 at 4:18
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.