A man pushes a 45 kg box from rest along a horizontal floor with a force of 87 N directed downward at 33 degrees to the horizontal, against a friction force of 62 N, over a distance of 13 m. How much work was done in moving the box?

Step-by-step solution

1. 1

   Keep only the part of the push that points along the motion. Work is $W=Fd\cos\theta$, so a force at $33^\circ$ to the floor contributes only its horizontal component:

   $F_{x}=87\cos 33^\circ=87(0.83867)=72.965\ \text{N}.$

   The vertical component $87\sin33^\circ=47.4\ \text{N}$ presses the box into the floor and does no work, because the box does not move vertically.

2. 2

   Note what the 45 kg mass is for. It is not needed for the work calculation — it would matter only if you were asked for the acceleration or the final speed. Problems often supply the mass as a distractor for part (a).

3. 3

   Find the net force along the direction of travel. Friction opposes motion with $62\ \text{N}$:

   $F_{\text{net}}=72.965-62=10.965\ \text{N}.$

   The box was pushed from rest and does move, so the push genuinely exceeds friction — a useful consistency check.

4. 4

   Multiply by the distance to get the net work.

   $W_{\text{net}}=F_{\text{net}}\,d=10.965\times 13=142.5\ \text{J}.$

   Rounding the component to $72.95$ before subtracting gives $142.4\ \text{J}$; carrying the full value gives $142.5\ \text{J}$, so round only at the end.

5. 5

   Separate the individual contributions if the question is read the other way. The push alone does $72.965\times 13=948.5\ \text{J}$, and friction does $-62\times 13=-806\ \text{J}$. Their sum, $948.5-806=142.5\ \text{J}$, is the same net work — and by the work-energy theorem it equals the kinetic energy the box gains.