Question

In Fig.7-47 , a cord runs around two massless, frictionless pulleys. A canister with mass m 20 kg hangs from one pulley, and you exert a force $\overrightarrow{\mathbf{F}}$on the free end of the cord. (a) What must be the magnitude of$\overrightarrow{\mathbf{F}}$ if you are to lift the canister at a constant speed? (b) To lift the canister by 2.0 cm , how far must you pull the free end of the cord? During that lift, what is the work done on the canister by (c) your force (via the cord) and (d) the gravitational force? (Hint: When a cord loops around a pulley as shown, it pulls on the pulley with a net force that is twice the tension in the cord.)

Short answer

(a) The magnitude of $\overrightarrow{F}$if the canister is lifted with constant speed is 98 N.

(b) The distance by the free end of the cord that should be pulled to lift the canister by 2.0 cm is 4.0 cm.

(c)The work done on the canister by the force applied via cord is 3.9 J .

(d) The work done on the canister by the gravitational force is $-3.9\mathrm{J}$.

Step-by-step solution

Given information

It is given that,

Mass of the canister is$m=20\mathrm{kg}$

The canister is lifted by the distance

$$\begin{gathered} d=2.0\mathrm{cm} \\ =0.02\mathrm{m} \end{gathered}$$

Determining the concept

The problem is based on Newton’s second law of motion which states that the rate of change of momentum of a body is equal in both magnitude and direction of the force acting on it. Also it deals with the work done which is the displacement of an object when force is applied on it. Find the magnitude of by applying Newton’s second law to the system. From the free body diagram, find the answer for part b. Then using the formula for work done, find the work done on the canister by the force applied via cord and by gravitational force.

Formulae:

$$\begin{gathered} F_{net}=ma \\ W=\overrightarrow{F},\overrightarrow{d} \end{gathered}$$

Where,

${\overrightarrow{F}}_{net}$ Is net force, dis displacement, mis the mass, a is an acceleration and Wis the work done.

(a) determining the magnitude of if the canister is lifted with a constant speed

Free body diagram:

The canister is moving with a constant speed, meaning the acceleration of the system is zero. It implies that the system is at equilibrium. Hence, Newton’s second law gives,

$F_{net}=0$

From above free body diagram,

$2T=mg$

And,

$F=T$

So,

role="math" localid="1657174594280" $$\begin{gathered} F=\frac{mg}{2} \\ =\frac{20\left(9.8\right)}{2} \\ F=98\mathrm{N} \end{gathered}$$

Therefore, the magnitude of$\overrightarrow{F}$ if the canister is lifted with a constant speed is 98 N

(b) Determining the distance by the free end of the cord that should be pulled to lift the canister by 2.0 cm

If the canister is lifted by 2.0 cm , the two cord segments will be shortened by the same amount. Therefore, the net distance through which the cord is pulled is 4.0 cm.

(c) Determining thework done on the canister by the force applied via cord

Now,$W=\overrightarrow{F}.\overrightarrow{d}$,

Since both $\overrightarrow{F}$ and d are in the downward direction, the work done by $\overrightarrow{F}$is,

$$\begin{gathered} W=F.d \\ W=\left(98\right)\left(0.04\right) \\ =3.9\mathrm{J} \end{gathered}$$

Therefore, the work done on the canister by the force applied via cord is $3.9\mathrm{J}$.

(d) Determining thework done on the canister by the gravitational force

Now,$W=\overrightarrow{F}.\overrightarrow{d}$,

Since the gravitational force$\overrightarrow{F_g}$ is in downward direction and$\overrightarrow{d}$ is in upward direction, the work done by the gravitational force is,

$$\begin{gathered} W=-F_g{d}_c \\ =-\left(20\right)\left(9.8\right)\left(0.020\right) \\ W=-3.9\mathrm{J} \end{gathered}$$

Therefore, the work done on the canister by the gravitational force is$-3.9\mathrm{J}$.