Question

Depicted in the plot above is the horizontal force a cube experience as it slides on a frictionless surface. The force that the cube experiences changes as a function of distance, which is shown on the horizontal axis.
What is the work done on the cube from $x=1m$ to $x=3m$?

• A. $3J$
• B. $7.25J$
• C. $8J$
• D. $4.25J$
• E. $7J$

Answer 1

Answer: (B)

$7.25J$

Area under F-x graph gives the work done by the force.

From figure, AD = 2 N BE =4 N CF =4.5 N DE = EF =1 s

Area of trap(DABE) $=\frac{1}{2}[AD+BE]\times DE=\frac{1}{2}\times [2+4]\times 1=3$ J

Area of trap(EBCF) $=\frac{1}{2}[CF+BE]\times EF=\frac{1}{2}\times [4.5+4]\times 1=4.25$ J

Thus work done on the cube from $x=1$ to $x=3$ m, $W=$ Area of trap(DABE) + Area of trap(EBCF)

$\therefore$ $W=3+4.25=7.25$ J