Question

(III) A grocery cart with mass of 16 kg is being pushed at constant speed up a 12° ramp by a force F_P which acts at an angle of 17° below the horizontal. Find the work done by each of the forces \(\left( {m\vec g,\;{{\vec F}_{\rm{N}}},\;{{\vec F}_{\rm{P}}}} \right)\) on the cart if the ramp is 7.5 m long.

Short answer

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

The work done by the normal force\(\left( {{F_{\rm{N}}}} \right)\)is\(0\;{\rm{J}}\).

The work done by the force \(\left( {{F_{\rm{P}}}} \right)\)is \(244\;{\rm{J}}\).

Step-by-step solution

Determination of work done

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.

Give information

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 \(d = 7.5\;{\rm{m}}\).

Calculate the work done by the applied force

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 \(244\;{\rm{J}}\).

Calculate the work done by the force of gravity

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 \( - 244\;{\rm{J}}\).

Calculate the work done by the normal force

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 \(0\;{\rm{J}}\).