Question

A work pushes a wheelbarrow with a horizontal force of $50 N,$ on level ground over a distance of$5.0 m$. If a friction force of $43 N$ acts on the wheelbarrow, in a direction opposite that of the worker, what work is done on the wheelbarrow by the worker?
$
A.\,{\text{250J}} \\
B.\,{\text{215J}} \\
C.\,35{\text{J}} \\
D.\,1{\text{0J}} \\
$

Answer 1

Hint: Follow the basic concepts of the work done by using its formula for the required solution. Like the work done can be expressed in various ways and the MKS unit of work done is Newton into metre, but we will use its SI unit, i.e. Joule as per the required solution.

Complete step by step answer:
Given that- Force applied is, $F = 50 N$
Distance moved or the displacement, $d = 5 m$
Work done on the wheelbarrow by the worker is $W = F.d$
Place the given known values in the above equation.
$W = 50 \times 5$
Simplify the above right hand side of the equation using basic mathematical operations
$W = 250 N.m$
Now, as per the conversion of the units from MKS system to SI
Newton into metre is equal to Joule. AS SI (System International Unit) of work is joule.
$\therefore W = 250J$
The required answer – the work done on the wheelbarrow by the worker is $250 J$.

Hence, from the given multiple choices – the option A is the correct answer.

Note: Always remember the units, derived units and SI units of the physical
quantities to get the relation between the two or more physical quantities. Always check the given units and the units asked in the solutions. All the quantities should have the same system of units. There are three types of the system of units.
-MKS System (Metre Kilogram Second)
-CGS System (Centimetre Gram Second)
-System International (SI)
Also, remember the conversional relations among the system of units to make all the given units in the same format.