Chapter 6, Problem 4P

A 1200-N crate rests on the floor. How much work is required to move it at constant speed (a) 5.0 m along the floor against a friction force of 230 N, and (b) 5.0 m vertically?

To determine

To Determine:(a) Work required moving it at a constant speed along the floor.

(b) Work required moving it at constant speed vertically.

Answer to Problem 4P

Solution: (a) 1150 J (b) 6000 J

Explanation of Solution

To move something against a frictional force, the force required will be equal or greater than frictional force, so the minimum force will be $F_k$ =230 N.

Given:

Weight of the crate, F= 1200 N.

Friction force, $F_k$ = 230 N.

Distance along the floor, $d_1=5.0\text{ m}$ .

Vertical Height, $d_2=5.0\text{ m}$

Formula used:

  $W=F\cdot d$

Where, F is the required force.

W is the work done.

dis the distance or height.

Calculation: (a) Work required moving it at a constant speed along the floor:

  $\begin{array}{c}W=F_k\cdot d_1\\ =230\times 5\\ =1150\text{J}\end{array}$

(b) Work required moving it at constant speed vertically

  $\begin{array}{c}W=F\times d_2\\ =1200\times 5\\ =6000\text{J}\end{array}$

Conclusion:Work required moving it at a constant speed along the floor is 1150 J and work required moving it at a constant speed along the floor 6000 J.