Chapter 6, Problem 10P

A 380-kg piano slides 2.9 m down a 25° incline and is kept from accelerating by a man who is pushing back on it parallel to the incline (Fig. 6-36). Determine: (a) the force exerted by the man, (b) the work done on the piano by the man, (c) the work done on the piano by the force of gravity, and (d) the net work done on the piano. Ignore friction.

Figure 6-36

To determine

The force exerted by the man.

Answer to Problem 10P

Solution:1574 N

Explanation of Solution

Given:

A 380-kg piano slides 2.9 m down a $25°$ incline and is kept from accelerating by a man who is pushing back on it parallel to the incline.

Formula used:

By Newton’s second law of motion:

  ${\displaystyle \sum F}=ma$

Where, m is the mass and a is the acceleration.

Calculation:

The various forces acting on the piano are shown in the figure.

  

The piano does not slide if applied force is $F=mg\sin \theta$

  $F=380\times 9.8\times \sin 25°$

  $\begin{array}{l}F=380\times 9.8\times 0.4224\\ F=1574\text{N}\end{array}$

Conclusion:The force exerted by the man is 1574 N.

To determine

The work done on the piano by the man.

Answer to Problem 10P

Solution:The work done on the piano by the man is $4565\text{J}$ .

Explanation of Solution

Given:A 380-kg piano slides 2.9 m down a $25°$ incline and is kept from accelerating by a man who is pushing back on it parallel to the incline.

Formula used:

The work done on the piano by the man can be obtained by:

  $W=F\cdot d$

Where, F is the applied force and d is the displacement.

Calculation:

The work done on the piano by the man,

  $\begin{array}{l}W=F\cdot d\\ W=\left(mg\sin \theta \right)d\end{array}$

  $\begin{array}{l}W=1574\times 2.9\\ W\simeq 4565\text{J}\end{array}$

Conclusion:The work done on the piano by the man is $4565\text{J}$ .

To determine

The work done on the piano by the force of gravity.

Answer to Problem 10P

Solution:The work done on the piano by the force of gravity is $-4565\text{J}$

Explanation of Solution

Given:A 380-kg piano slides 2.9 m down a $25°$ incline and is kept from accelerating by a man who is pushing back on it parallel to the incline.

Formula used:

The work done on the piano by force of gravity can be obtained by:

  $W=F\cdot d$

Where, F is the applied force and d is the displacement.

Calculation:

The work done on the piano by the force of gravity is

  $W=-Fd=-\left(mg\sin \theta \right)d$

  $\begin{array}{l}W=-1574\times 2.9\\ W\simeq -4565\text{J}\end{array}$

Conclusion:The work done on the piano by the force of gravity is $-4565\text{J}$ .

To determine

The net workdone on the piano.

Answer to Problem 10P

Solution:The net work done on the piano is zero.

Explanation of Solution

The various forces acting on the piano are shown in the figure.

  

Given:A 380-kg piano slides 2.9 m down a $25°$ incline and is kept from accelerating by a man who is pushing back on it parallel to the incline.

Formula used:

The work done on the piano by force of gravity can be obtained by:

  $W=F\cdot d$

Where, F is the applied force and d is the displacement.

Calculation:

Since the net displacement of the piano is zero, the net work done on the piano is zero.

Conclusion:Thenet work done on the piano is zero.