Question

A box of mass 4.0 kg is accelerated from rest by a force across a floor at a rate of \(2.0\;{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right.} {{{\rm{s}}^{\rm{2}}}}}\) for 7.0 s. Find the net work done on the box.

Short answer

The net work done on the box is 392 J.

Step-by-step solution

Understanding of displacement

The net displacement can be defined as the vector addition of a particle's individual displacements in various directions. If the particle's initial and final points are similar, then its net displacement value will be zero.

Given information

Given data:

The mass of the box is\(m = 4.0\;{\rm{kg}}\).

The acceleration of the box is \(a = 2.0\;{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}}{{{\rm{s}}^{\rm{2}}}}}} \right.} {{{\rm{s}}^{\rm{2}}}}}\).

The time is\(t = 7.0\;{\rm{s}}\).

It is given that the box starts from rest.Therefore, the initial velocity of the box will be \(u = 0\).

Calculate the displacement of the box

The displacement of the box can be calculated as:

\(\begin{aligned}{}s &= ut + \frac{1}{2}a{t^2}\\s &= \left( 0 \right)\left( {7.0\;{\rm{s}}} \right) + \frac{1}{2}\left( {2.0\;{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right.} {{{\rm{s}}^{\rm{2}}}}}} \right){\left( {7.0\;{\rm{s}}} \right)^2}\\s& = 49\;{\rm{m}}\end{aligned}\)

Calculate the work done on the box

The expression for the force is given by:

\(F = ma\)

The work done on the box due to this force can be calculated as:

\(\begin{aligned}{}W &= Fs\\W &= mas\\W &= \left( {4.0\;{\rm{kg}}} \right)\left( {2.0\;{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right.} {{{\rm{s}}^{\rm{2}}}}}} \right)\left( {49\;{\rm{m}}} \right)\\W &= 392\;{\rm{J}}\end{aligned}\)

Thus, the required work done on the box is\(392\;{\rm{J}}\).