Question

(II) 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 Problem 10.

Short answer

(a) The force exerted by the man is \(1574\;{\rm{N}}\).

(b) The work done on the piano by the man is \( - 4564.10\;{\rm{J}}\).

(c) The work done on the piano by the force of gravity is \(4564.10\;{\rm{J}}\).

(d) The net work done on the piano is \(0\;{\rm{J}}\).

Step-by-step solution

Determination of force value

The value of the force can be obtained by evaluating the value of the work done and the displacement of the object.

Given information

The mass of the piano is \(m = 380\;{\rm{kg}}\).

The displacement of the piano is \(d = 2.9\;{\rm{m}}\).

The angle of inclination is \(\theta = 25^\circ \).

Calculate the force exerted by the man

(a)

Draw a free body diagram.

Here, \(m\) is the mass of the piano, \(g\) is the acceleration due to gravity, \({F_{\rm{P}}}\) is the force exerted by the man, and \({F_{\rm{N}}}\) is the normal force.

Apply the equilibrium condition along the horizontal direction.

\(\begin{aligned}\sum {F_{\rm{x}}} &= 0\\{F_{\rm{P}}} - mg\sin \theta &= 0\\{F_{\rm{P}}} &= mg\sin \theta \end{aligned}\)

Substitute the values in the above expression.

\(\begin{aligned}{}{F_{\rm{P}}} &= \left( {380\;{\rm{kg}}} \right)\left( {9.8\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right.}{{{\rm{s}}^{\rm{2}}}}}} \right)\sin \left( {25^\circ } \right)\\{F_{\rm{P}}} &= 1573.83{\rm{ N}}\\{F_{\rm{P}}} \approx 1574\;{\rm{N}}\end{aligned}\)

Thus, the force exerted by the man is \(1574\;{\rm{N}}\).

Calculate the work done on the piano by the man

(b)

The work done by the man on the piano can be calculated as:

\(\begin{aligned}{W_{\rm{P}}} &= {F_P}d\cos \left( {180^\circ } \right)\\{W_{\rm{P}}} &= \left( {1573.83\;{\rm{N}}} \right)\left( {2.9\;{\rm{m}}} \right)\cos \left( {180^\circ } \right)\\{W_{\rm{P}}} &= - 4564.10\end{aligned}\)

Thus, the work done on the piano by the man is \( - 4564.10\;{\rm{J}}\).

Calculate the work done on the piano by the force of gravity

(c)

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

\(\begin{aligned}{W_{\rm{G}}} &= {F_{\rm{G}}}s\cos \left( {90^\circ - 25^\circ } \right)\\{W_{\rm{G}}} &= mgs\cos \left( {65^\circ } \right)\\{W_{\rm{G}}} &= \left( {380\;{\rm{kg}}} \right)\left( {9.8\;{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right.} {{{\rm{s}}^{\rm{2}}}}}} \right)\left( {2.9\;{\rm{m}}} \right)\cos \left( {65^\circ } \right)\\{W_{\rm{G}}} &= 4564.10\;{\rm{J}}\end{aligned}\)

Thus, the work done on the piano by the force of gravity is \(4564.10\;{\rm{J}}\).

Calculate the net work done on the piano

(d)

The net work done on the piano can be calculated as:

\(\begin{aligned}{W_{{\rm{net}}}} &= {W_{\rm{P}}} + {W_{\rm{G}}}\\{W_{{\rm{net}}}} &= \left( { - 4564.10\;{\rm{J}}} \right) + \left( {4564.10\;{\rm{J}}} \right)\\{W_{{\rm{net}}}} &= 0\;{\rm{J}}\end{aligned}\)

Thus, the net work done on the piano is \(0\;{\rm{J}}\).