91Ó°ÊÓ

The value of the work done can be calculated by multiplying the magnitude of the force by the magnitude of the displacement of the object.

Given data:

The mass of the cart is\(m = 16\;{\rm{kg}}\).

The force\({F_{\rm{P}}}\)acting on the cart at an angle of\(\phi = 17^\circ \).

The inclination of the ramp is\(\theta = 12^\circ \).

The length of the ramp is \(d = 7.5\;{\rm{m}}\).

Draw a free body diagram.

Here,\({F_{\rm{N}}}\)is the normal force and\({F_{\rm{P}}}\)is the applied force.

Apply the equilibrium condition along the horizontal direction.

\(\begin{aligned}{}\sum {F_{\rm{x}}} = 0\\{F_{\rm{P}}}\cos \left( {\phi + \theta } \right) - mg\sin \theta = 0\\{F_{\rm{P}}}\cos \left( {\phi + \theta } \right) = mg\sin \theta \\{F_{\rm{P}}} = \frac{{mg\sin \theta }}{{\cos \left( {\phi + \theta } \right)}}\end{aligned}\)

Substitute the values in the above equation.

\(\begin{aligned}{}{F_{\rm{P}}} = \frac{{\left( {16\;{\rm{kg}}} \right)\left( {9.8\;{{\rm{m}} \mathord {\left/{\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right.} {{{\rm{s}}^{\rm{2}}}}}} \right)\sin \left( {12^\circ } \right)}}{{\cos \left( {17^\circ + 12^\circ } \right)}}\\{F_{\rm{P}}} = 37.27\;{\rm{N}}\end{aligned}\)

The work done by the force\({F_{\rm{P}}}\)can be calculated as:

\(\begin{aligned}{}{W_{{{\rm{F}}_{\rm{P}}}}} = {F_{\rm{P}}}d\cos \left( {\phi + \theta } \right)\\{W_{{{\rm{F}}_{\rm{P}}}}} = \left( {37.27\;{\rm{N}}} \right)\left( {7.5\;{\rm{m}}} \right)\cos \left( {17^\circ + 12^\circ } \right)\\{W_{{{\rm{F}}_{\rm{P}}}}} = 244.47{\rm{ J}}\\{W_{{{\rm{F}}_{\rm{P}}}}} \approx 244\;{\rm{J}}\end{aligned}\)

Thus, the work done by the force \({F_{\rm{P}}}\)is \(244\;{\rm{J}}\).

The work done by the gravitational force\(\left( {mg} \right)\)can be calculated as:

\(\begin{aligned}{W_{{\rm{mg}}}} + {W_{{{\rm{F}}_{\rm{P}}}}} &= 0\\{W_{{\rm{mg}}}} &= - {W_{{{\rm{F}}_{\rm{P}}}}}\\{W_{{\rm{mg}}}} &= - 244\;{\rm{J}}\end{aligned}\)

Thus, the work done by the gravitational force \(\left( {mg} \right)\)is \( - 244\;{\rm{J}}\).

The work done by the normal force will be zero as the force is perpendicular to the displacement. Mathematically, it is calculated as:

\(\begin{aligned}{W_{\rm{N}}} &= {F_{\rm{N}}}d\cos \left( {90^\circ } \right)\\{W_{\rm{N}}} &= 0\;{\rm{J}}\end{aligned}\)

Thus, the work done by the normal force is \(0\;{\rm{J}}\).