Question

raising a block at constant speed. the work done to overcome gravity. what is external work done by f2? a. $mgsin\theta d$ b. $mgcos\theta d$ c. $mgd$ d. $mgh$ e. a and d

Explanation:

Step1: Recall work - formula

The work - done formula is $W = F\cdot d\cdot\cos\theta$, where $F$ is the force, $d$ is the displacement, and $\theta$ is the angle between the force and the displacement.

Step2: Analyze the force $F_2$ and displacement

The force $F_2$ is used to lift the block of mass $M$ vertically through a height $h$. The displacement of the block in the direction of $F_2$ is $h$. Also, when considering the motion along the inclined - plane of length $d$, the vertical height $h = d\sin\theta$. The force $F_2$ balances the gravitational force $Mg$ (since the block is moving at a constant speed). The work done by $F_2$ is $W=F_2\cdot h$. Since $F_2 = Mg$, the work done $W = Mg\cdot h$. Also, since $h = d\sin\theta$, the work done $W=Mg\sin\theta d$.

Answer:

E. A and D