Question

raising a block at constant speed. the work done to overcome gravity. what is external work done by f1? a. mg sinθd b. mg cosθd c. mg d d. mg h e. a and d

Explanation:

Step1: Recall work - formula

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

Step2: Analyze forces on the block on the incline

Since the block is moving at a constant speed, the net force on the block is zero. The force $F_1$ that moves the block up the incline must balance the component of the gravitational force along the incline. The component of the gravitational force along the incline is $Mg\sin\theta$. So, $F_1 = Mg\sin\theta$.

Step3: Calculate work - done by $F_1$

The displacement along the incline is $d$. Using the work - done formula $W = F_1d$, substituting $F_1 = Mg\sin\theta$, we get $W = Mg\sin\theta d$. Also, from the geometry of the incline, $h = d\sin\theta$, so $W = Mg h$.

Answer:

E. A and D